Antiderivatives(III) | NEB Class 11 Mathematics Notes

Chapter Overview

In our previous study of antiderivatives, we focused on finding a general function (with a constant +c) whose derivative matches our integrand. In this chapter, we transition to Definite Integrals, where we find a specific numerical value by evaluating the antiderivative between two specific points called limits.

This is one of the most critical chapters in Calculus. It forms the foundation for finding areas, volumes, and solving physics problems involving work and energy. In the NEB Board Exams, this section carries significant weight in both short and long answer questions.

What this chapter covers

Definition and Evaluation of Definite IntegralsThe Fundamental Theorem of CalculusProperties of Definite IntegralsChange of Limits in Substitution MethodIntegration by Parts for Definite IntegralsIntegrals Involving Modulus FunctionsEven and Odd Function PropertiesDummy Variable PropertySplitting Integrals Using Additive Property

Learning Objectives

1

Define and evaluate definite integrals using the Fundamental Theorem of Calculus correctly

2

Apply properties of definite integrals to simplify complex problems efficiently

3

Use the method of substitution with proper change of limits

4

Apply Integration by Parts to definite integrals with confidence

5

Evaluate integrals involving modulus functions by splitting intervals at critical points

6

Distinguish between even and odd functions and apply symmetry properties

7

Avoid common mistakes like leaving '+c' in definite integral answers or forgetting to change substitution limits

Definite Integral = Net Area = A Single Number

Key Insight

The fundamental difference between indefinite and definite integrals is that an indefinite integral gives you a family of functions (F(x) + c), while a definite integral gives you a single numerical value.

When you write $$\int_{a}^{b} f(x) \, dx = F(b) - F(a)$$ you are computing the net signed area under the curve from x = a to x = b. The result is a pure number — no variable, no constant of integration.

This number represents the total accumulation of the quantity f(x) over the interval [a, b].

In NEB exams: The moment you see $$\int_{a}^{b}$$ with upper and lower limits, you know the answer must be a number. If you see '+c' in a definite integral answer, it's automatically wrong.

Key Definitions

Definite Integral

If F(x) is an antiderivative of f(x), the difference F(b) − F(a) is the definite integral of f(x) from a to b. It represents the net signed area under the curve.

$$\int_{a}^{b} f(x) \, dx = [F(x)]_{a}^{b} = F(b) - F(a)$$

Lower Limit of Integration

In the notation $$\int_{a}^{b} f(x) \, dx$$, the value 'a' is the lower limit. It is the starting point of the interval where integration begins.

Upper Limit of Integration

In the notation $$\int_{a}^{b} f(x) \, dx$$, the value 'b' is the upper limit. It is the ending point of the interval where integration stops.

Integrand

The function f(x) being integrated. It is the expression that appears between the integral sign and the differential (dx).

Fundamental Theorem of Calculus

This theorem bridges differentiation and integration. It states that if F is an antiderivative of f on [a, b], then $$\int_{a}^{b} f(x) \, dx = F(b) - F(a)$$.

Even Function

A function where f(−x) = f(x) for all x. The graph is symmetric about the y-axis.

$$\int_{-a}^{a} f(x) \, dx = 2\int_{0}^{a} f(x) \, dx \quad \text{(if f is even)}$$

Odd Function

A function where f(−x) = −f(x) for all x. The graph has rotational symmetry about the origin.

$$\int_{-a}^{a} f(x) \, dx = 0 \quad \text{(if f is odd)}$$

The Fundamental Theorem of Calculus

This theorem bridges the gap between differentiation and integration. It tells us that if you integrate a rate of change, you get the net change in the original function.

$$\int_{a}^{b} f(x) \, dx = [F(x)]_{a}^{b} = F(b) - F(a)$$

Why it works (Derivation):

1Let G(x) be defined as $$G(x) = \int_{a}^{x} f(t) \, dt$$. This represents the accumulated area from a to x.

2Let F(x) be any other antiderivative of f(x). Since F and G are both antiderivatives of the same function f, they differ only by a constant c:

3$$F(x) = G(x) + c$$

4To find c, substitute x = a: $$F(a) = G(a) + c$$. Since $$G(a) = \int_{a}^{a} f(t) \, dt = 0$$, we get c = F(a).

5Now substitute x = b: $$F(b) = G(b) + c = \int_{a}^{b} f(t) \, dt + F(a)$$

6Rearranging: $$\int_{a}^{b} f(t) \, dt = F(b) - F(a)$$. Since the variable of integration is a dummy variable, we can replace t with x.

Example: Evaluate $$\int_{1}^{3} x^{2} \, dx$$.

Step 1: Find the antiderivative: $$\int x^{2} \, dx = \frac{x^{3}}{3}$$.

Step 2: Apply limits: $$\left[\frac{x^{3}}{3}\right]_{1}^{3} = \frac{3^{3}}{3} - \frac{1^{3}}{3} = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}$$.

When to use: Use the Fundamental Theorem directly for polynomials, trigonometric functions, and exponential functions when the antiderivative is known or easily found.

Change of Limits in Substitution

When you substitute u = g(x) in a definite integral, the limits a and b (which are x-values) must also be converted to the corresponding u-values. This is a very common area where students lose marks.

Example: Suppose we substitute u = 1 + 3x³ in an integral from x = 0 to x = 1.

When x = 0: u = 1 + 3(0)³ = 1 (new lower limit)
When x = 1: u = 1 + 3(1)³ = 4 (new upper limit)

The integral becomes a u-integral from 1 to 4 — no need to convert back to x!

When to use: Always change limits when using substitution in definite integrals. If you forget, you must substitute back to x before applying the original limits — which wastes time and risks errors.

Complete Formula Sheet with Derivations

Fundamental Theorem of Calculus (Power Rule Application)

\int_{a}^{b} x^{n} \, dx = \left[\frac{x^{n+1}}{n+1}\right]_{a}^{b} = \frac{b^{n+1} - a^{n+1}}{n+1} \quad (n \neq -1)

This is the most basic formula. Apply the power rule for integration, then evaluate at the upper and lower limits.

1The antiderivative of xⁿ is xⁿ⁺¹/(n+1) for n ≠ −1.

2Evaluate at upper limit b: $$\frac{b^{n+1}}{n+1}$$

3Evaluate at lower limit a: $$\frac{a^{n+1}}{n+1}$$

4Subtract: $$\frac{b^{n+1}}{n+1} - \frac{a^{n+1}}{n+1} = \frac{b^{n+1} - a^{n+1}}{n+1}$$

Property of Identical Limits (Zero Property)

\int_{a}^{a} f(x) \, dx = 0

When upper and lower limits are the same, there is no interval — so the area is zero.

1By the Fundamental Theorem: $$\int_{a}^{a} f(x) \, dx = F(a) - F(a) = 0$$

2Geometrically: the 'width' of the region is zero, so the area must be zero.

Swapping Limits Property

\int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx

Reversing the limits changes the sign of the integral. This is useful when you need to manipulate integral expressions.

1$$\int_{a}^{b} f(x) \, dx = F(b) - F(a)$$

2$$\int_{b}^{a} f(x) \, dx = F(a) - F(b) = -(F(b) - F(a))$$

3Therefore: $$\int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx$$

Additive Property (Interval Splitting)

\int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx \quad (a < c < b)

Essential for modulus functions and piecewise functions. You can split the interval at any point c between a and b.

1$$\int_{a}^{c} f(x) \, dx = F(c) - F(a)$$

2$$\int_{c}^{b} f(x) \, dx = F(b) - F(c)$$

3Adding: $$[F(c) - F(a)] + [F(b) - F(c)] = F(b) - F(a) = \int_{a}^{b} f(x) \, dx$$

4The F(c) terms cancel out.

Dummy Variable Property

\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(t) \, dt = \int_{a}^{b} f(\theta) \, d\theta

The variable of integration is called a 'dummy variable' because it does not appear in the final answer. The result depends only on the function and the limits.

1All three integrals equal F(b) − F(a), regardless of whether we call the variable x, t, or θ.

2The definite integral is a number that depends on the function f and the limits a, b — not on the symbol used for the variable.

Even Function Property

\int_{-a}^{a} f(x) \, dx = 2\int_{0}^{a} f(x) \, dx \quad \text{if } f(-x) = f(x)

Applies to functions like x², cos x, |x|. The graph is symmetric about the y-axis, so the area from −a to 0 equals the area from 0 to a.

1Split: $$\int_{-a}^{a} f(x) \, dx = \int_{-a}^{0} f(x) \, dx + \int_{0}^{a} f(x) \, dx$$

2In the first integral, substitute x = −t, dx = −dt. When x = −a, t = a; when x = 0, t = 0:

3$$\int_{-a}^{0} f(x) \, dx = \int_{a}^{0} f(-t)(-dt) = \int_{0}^{a} f(-t) \, dt = \int_{0}^{a} f(t) \, dt$$

4(since f is even, f(−t) = f(t))

5Therefore: $$\int_{-a}^{a} f(x) \, dx = \int_{0}^{a} f(x) \, dx + \int_{0}^{a} f(x) \, dx = 2\int_{0}^{a} f(x) \, dx$$

Odd Function Property

\int_{-a}^{a} f(x) \, dx = 0 \quad \text{if } f(-x) = -f(x)

Applies to functions like x, x³, sin x. The areas on the positive and negative sides cancel each other out perfectly.

1Split: $$\int_{-a}^{a} f(x) \, dx = \int_{-a}^{0} f(x) \, dx + \int_{0}^{a} f(x) \, dx$$

2Substitute x = −t in the first integral: $$\int_{a}^{0} f(-t)(-dt) = \int_{0}^{a} f(-t) \, dt = \int_{0}^{a} [-f(t)] \, dt = -\int_{0}^{a} f(t) \, dt$$

3(since f is odd, f(−t) = −f(t))

4Therefore: $$\int_{-a}^{a} f(x) \, dx = -\int_{0}^{a} f(x) \, dx + \int_{0}^{a} f(x) \, dx = 0$$

Even and Odd Functions — Symmetry Shortcut

Recognizing even and odd functions can save enormous time in exams. Instead of computing the full integral, you can use symmetry properties to get the answer instantly.

$$\text{Even: } f(-x) = f(x) \Rightarrow \int_{-a}^{a} f(x) \, dx = 2\int_{0}^{a} f(x) \, dx \newline \text{Odd: } f(-x) = -f(x) \Rightarrow \int_{-a}^{a} f(x) \, dx = 0$$

Example: $$\int_{-2}^{2} x^{3} \, dx = 0$$ instantly, because x³ is odd.

$$\int_{-\pi/2}^{\pi/2} \cos x \, dx = 2\int_{0}^{\pi/2} \cos x \, dx = 2[\sin x]_{0}^{\pi/2} = 2(1) = 2$$, because cos x is even.

When to use: Always check if the integrand is even or odd when the limits are symmetric about 0 (i.e., from −a to a). This is especially common in MCQs where the answer is 0.

Memory Tricks — Never Forget These

Trick: Three quick memory hooks that will save you marks in NEB exams for definite integrals.

Steps to Remember:

'Top minus Bottom': Always remember F(upper limit) − F(lower limit). Write the upper limit first, then subtract the lower limit. A common error is reversing this — which gives the negative of the correct answer.
'Odd is Zero': For limits like −a to a, if the function is odd (like x, x³, sin x, tan x), the answer is immediately 0. This saves huge time in MCQs! Just verify f(−x) = −f(x) and write the answer.
'Dummy Variable': The name of the variable (x, t, θ) doesn't change the value of the definite integral. So $$\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(t) \, dt$$. Don't get confused if the variable changes in a substitution problem.

Mnemonic: Top minus Bottom for limits | Odd is Zero for symmetry | Dummy variable name doesn't matter

Common Mistakes

These mistakes cause the most mark losses in NEB exams for definite integrals.

Forgetting to change limits in substitution: When you substitute u = g(x), students often keep the x-limits for the u-variable. Always convert: when x = a, find u = g(a); when x = b, find u = g(b).
Sign errors in limits: Subtracting F(a) − F(b) instead of F(b) − F(a). Remember: Upper limit first, lower limit second.
Ignoring Modulus: Integrating |x| directly as x²/2 without splitting the interval. You must find where the inside function equals zero and split the integral at that point.
Leaving '+c' in definite integrals: Definite integrals result in a number; never add the constant of integration 'c' in the final answer. The constant cancels out when you compute F(b) − F(a).

NEB Exam Traps — These Appear Every Year

NEB Exam Warning

Board examiners specifically test these edge cases in definite integrals. Knowing them can save 2–5 marks.

Questions with symmetric limits (−a to a) that test even/odd properties — always check symmetry first before doing full calculations.
Substitution questions where forgetting to change limits leads to wrong answers. If you see a composite function inside the integral (like √(1+3x³)), substitution is likely needed.
Modulus integral questions (like |2x−1| or |x²−1|) — you must split at the root where the inside expression equals zero. This is a favorite 4-mark question.
Integration by Parts combined with definite limits — don't forget to apply limits at every step, not just at the end.
Questions using the property $$\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$$ for trigonometric integrals — this is essential for fractions like $$\frac{\sin x}{\sin x + \cos x}$$.

Question Recognition Guide

Keyword / Phrase in Question	Use This
Standard polynomial or simple trig function with limits	Use Fundamental Theorem directly: F(b) − F(a)
Derivative of a part of the function is present (e.g., ln x and 1/x)	Use substitution and remember to change limits
Integrand contains \|f(x)\| (modulus/absolute value)	Find root where f(x)=0, split integral at that point
Product of functions like x·sin x, x·eˣ, or x·ln x	Integration by Parts with definite limits
Symmetric limits (−a to a) with even/odd function	Apply symmetry: Even → 2× half, Odd → 0
Trigonometric fraction like sin x/(sin x + cos x)	Use property: ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a−x)dx, then add
√(a² − x²) form with limits 0 to a	Substitute x = a sin θ, change limits to 0 to π/2
Identical upper and lower limits	Answer is immediately 0 (zero property)

Solved Examples (Easy → Hard)

1

Evaluate $$\int_{1}^{4} \frac{1}{\sqrt{x}} \, dx$$.

Easy

Solution

Step 1: Rewrite the integrand using exponent rules:
$$\int_{1}^{4} x^{-1/2} \, dx$$

Step 2: Apply the power rule for integration:
$$\left[\frac{x^{-1/2+1}}{-1/2+1}\right]_{1}^{4} = \left[\frac{x^{1/2}}{1/2}\right]_{1}^{4} = [2\sqrt{x}]_{1}^{4}$$

Step 3: Apply the limits (upper minus lower):
$$2\sqrt{4} - 2\sqrt{1} = 2(2) - 2(1) = 4 - 2 = \boxed{2}$$

∴ 2

2

Evaluate $$\int_{0}^{\pi/4} \tan^{2} x \, dx$$.

Medium

Solution

Step 1: Use the trigonometric identity $$\tan^{2} x = \sec^{2} x - 1$$.

$$\int_{0}^{\pi/4} (\sec^{2} x - 1) \, dx$$

Step 2: Integrate term by term:
$$[\tan x - x]_{0}^{\pi/4}$$

Step 3: Apply the limits:
$$(\tan \frac{\pi}{4} - \frac{\pi}{4}) - (\tan 0 - 0) = \left(1 - \frac{\pi}{4}\right) - (0 - 0) = \boxed{1 - \frac{\pi}{4}}$$

∴ 1 − π/4

3

Evaluate $$\int_{0}^{1} x^{2} \sqrt{1+3x^{3}} \, dx$$ using substitution.

Hard

Solution

Step 1 (Substitution): Let $$u = 1 + 3x^{3}$$. Then $$du = 9x^{2} \, dx \Rightarrow \frac{du}{9} = x^{2} \, dx$$.

Step 2 (Change Limits):
When x = 0: $$u = 1 + 3(0)^{3} = 1$$ (new lower limit)
When x = 1: $$u = 1 + 3(1)^{3} = 4$$ (new upper limit)

Step 3 (New Integral):
$$\int_{1}^{4} \sqrt{u} \cdot \frac{du}{9} = \frac{1}{9} \int_{1}^{4} u^{1/2} \, du$$

Step 4 (Integrate):
$$\frac{1}{9} \left[\frac{u^{3/2}}{3/2}\right]_{1}^{4} = \frac{1}{9} \cdot \frac{2}{3} [u^{3/2}]_{1}^{4} = \frac{2}{27}(4^{3/2} - 1^{3/2})$$

Step 5 (Final Value):
$$\frac{2}{27}(8 - 1) = \frac{2}{27} \times 7 = \boxed{\frac{14}{27}}$$

∴ 14/27

MCQ Practice

Mixed Level — MCQ

1

The value of $$\int_{a}^{a} f(x) \, dx$$ is:

Correct!

Incorrect. Correct: 0

Solution

By the property of identical limits: $$\int_{a}^{a} f(x) \, dx = F(a) - F(a) = \boxed{0}$$

This is because there is no interval to integrate over when both limits are the same.

∴ 0

2

$$\int_{-1}^{1} x^{3} \, dx$$ equals:

Correct!

Incorrect. Correct: 0

Solution

Check if f(x) = x³ is odd or even:
f(−x) = (−x)³ = −x³ = −f(x)

Since f(x) is odd and the limits are symmetric (−1 to 1):
$$\int_{-1}^{1} x^{3} \, dx = \boxed{0}$$

The areas on the negative and positive sides are equal in magnitude but opposite in sign, so they cancel perfectly.

∴ 0

3

If we substitute $$u = \ln x$$ in $$\int_{1}^{e} \frac{1}{x} \, dx$$, the new limits are:

Correct!

Incorrect. Correct: 0 to 1

Solution

Convert the limits using u = ln x:

When x = 1: $$u = \ln 1 = \textbf{0}$$ (new lower limit)
When x = e: $$u = \ln e = \textbf{1}$$ (new upper limit)

Also, $$du = \frac{1}{x} dx$$, so the integral becomes:
$$\int_{0}^{1} du = [u]_{0}^{1} = 1 - 0 = \boxed{1}$$

∴ 0 to 1

4

The value of $$\int_{0}^{\pi/2} \cos x \, dx$$ is:

Correct!

Incorrect. Correct: 1

Solution

Antiderivative of cos x is sin x.

$$\int_{0}^{\pi/2} \cos x \, dx = [\sin x]_{0}^{\pi/2} = \sin\frac{\pi}{2} - \sin 0 = 1 - 0 = \boxed{1}$$

∴ 1

5

Which property is correct?

Correct!

Incorrect. Correct: Both B and C

Solution

Property B (Swapping Limits): $$\int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx$$ — This is always true.

Property C: $$\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$$ — This is true for intervals starting at 0.

Both B and C are standard properties. The answer is Both B and C.

∴ Both B and C

6

Evaluate $$\int_{0}^{2} e^{x} \, dx$$:

Correct!

Incorrect. Correct: e² − 1

Solution

Antiderivative of eˣ is eˣ.

$$\int_{0}^{2} e^{x} \, dx = [e^{x}]_{0}^{2} = e^{2} - e^{0} = e^{2} - 1 = \boxed{e^{2} - 1}$$

Remember: e⁰ = 1, not 0!

∴ e² − 1

7

$$\int_{0}^{1} \frac{1}{1+x^{2}} \, dx$$ is:

Correct!

Incorrect. Correct: π/4

Solution

Antiderivative of $$\frac{1}{1+x^{2}}$$ is $$\tan^{-1} x$$.

$$\int_{0}^{1} \frac{1}{1+x^{2}} \, dx = [\tan^{-1} x]_{0}^{1} = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \boxed{\frac{\pi}{4}}$$

∴ π/4

8

The definite integral of an even function f(x) from −a to a is:

Correct!

Incorrect. Correct: $$2\int_{0}^{a} f(x) dx$$

Solution

For an even function where f(−x) = f(x):

$$\int_{-a}^{a} f(x) \, dx = 2\int_{0}^{a} f(x) \, dx$$

This is because the graph is symmetric about the y-axis, so the area on the left equals the area on the right.

∴ 2∫₀ᵃ f(x)dx

9

To evaluate $$\int_{0}^{1} |2x-1| \, dx$$, we split the limit at:

Correct!

Incorrect. Correct: x = 1/2

Solution

For modulus integrals, find where the inside expression equals zero:
2x − 1 = 0 → x = 1/2

Split the integral:
$$\int_{0}^{1/2} -(2x-1) \, dx + \int_{1/2}^{1} (2x-1) \, dx$$

On [0, 1/2]: 2x−1 is negative, so |2x−1| = −(2x−1)
On [1/2, 1]: 2x−1 is positive, so |2x−1| = 2x−1

∴ x = 1/2

10

$$\int_{1}^{2} 3 \, dx$$ equals:

Correct!

Incorrect. Correct: 3

Solution

Antiderivative of the constant 3 is 3x.

$$\int_{1}^{2} 3 \, dx = [3x]_{1}^{2} = 3(2) - 3(1) = 6 - 3 = \boxed{3}$$

Geometrically: this is a rectangle with height 3 and width (2−1) = 1, so area = 3 × 1 = 3.

∴ 3

Short Questions (2 Marks Each)

1

Evaluate $$\int_{0}^{3} (4x-1) \, dx$$.

[2 marks]

Solution

Find the antiderivative:
$$\int (4x-1) \, dx = \frac{4x^{2}}{2} - x = 2x^{2} - x$$

Apply limits:
$$[2x^{2} - x]_{0}^{3} = (2 \times 9 - 3) - (0 - 0) = 18 - 3 = \boxed{15}$$

∴ 15

2

Evaluate $$\int_{0}^{\pi} \sin x \, dx$$.

[2 marks]

Solution

Antiderivative of sin x is −cos x.

$$[-\cos x]_{0}^{\pi} = -\cos \pi - (-\cos 0) = -(-1) + 1 = 1 + 1 = \boxed{2}$$

∴ 2

3

Use properties to find $$\int_{-2}^{2} x^{5} \, dx$$.

[2 marks]

Solution

Check symmetry: f(x) = x⁵
f(−x) = (−x)⁵ = −x⁵ = −f(x)

Since f(x) = x⁵ is an odd function and the limits are symmetric (−2 to 2):
$$\int_{-2}^{2} x^{5} \, dx = \boxed{0}$$

∴ 0

4

Evaluate $$\int_{1}^{e} \frac{1}{x} \, dx$$.

[2 marks]

Solution

Antiderivative of 1/x is ln|x|.

$$[\ln|x|]_{1}^{e} = \ln e - \ln 1 = 1 - 0 = \boxed{1}$$

∴ 1

5

Find $$\int_{0}^{1} e^{2x} \, dx$$.

[2 marks]

Solution

Antiderivative of e^(2x) is e^(2x)/2.

$$\left[\frac{e^{2x}}{2}\right]_{0}^{1} = \frac{e^{2}}{2} - \frac{e^{0}}{2} = \frac{e^{2} - 1}{2} = \boxed{\frac{e^{2}-1}{2}}$$

∴ (e² − 1)/2

6

Evaluate $$\int_{4}^{9} \sqrt{x} \, dx$$.

[2 marks]

Solution

Rewrite: $$\sqrt{x} = x^{1/2}$$

Antiderivative: $$\frac{2}{3}x^{3/2}$$

$$\left[\frac{2}{3}x^{3/2}\right]_{4}^{9} = \frac{2}{3}(9^{3/2} - 4^{3/2}) = \frac{2}{3}(27 - 8) = \frac{2}{3} \times 19 = \boxed{\frac{38}{3}}$$

∴ 38/3

7

If $$\int_{0}^{a} 3x^{2} \, dx = 8$$, find a.

[2 marks]

Solution

$$\int_{0}^{a} 3x^{2} \, dx = [x^{3}]_{0}^{a} = a^{3} - 0 = a^{3}$$

Given a³ = 8:
$$a = \sqrt[3]{8} = \boxed{2}$$

∴ a = 2

8

Evaluate $$\int_{0}^{\pi/2} \cos^{2} x \, dx$$.

[2 marks]

Solution

Use identity: $$\cos^{2} x = \frac{1 + \cos 2x}{2}$$

$$\int_{0}^{\pi/2} \frac{1 + \cos 2x}{2} \, dx = \frac{1}{2}\left[x + \frac{\sin 2x}{2}\right]_{0}^{\pi/2}$$

$$= \frac{1}{2}\left[\left(\frac{\pi}{2} + \frac{\sin \pi}{2}\right) - \left(0 + \frac{\sin 0}{2}\right)\right] = \frac{1}{2} \times \frac{\pi}{2} = \boxed{\frac{\pi}{4}}$$

∴ π/4

9

Evaluate $$\int_{0}^{1} \frac{x}{x^{2}+1} \, dx$$.

[2 marks]

Solution

Substitution: Let u = x² + 1, then du = 2x dx, so x dx = du/2.

Change limits: x = 0 → u = 1; x = 1 → u = 2.

$$\int_{1}^{2} \frac{1}{2u} \, du = \frac{1}{2}[\ln u]_{1}^{2} = \frac{1}{2}(\ln 2 - \ln 1) = \boxed{\frac{1}{2} \ln 2}$$

∴ (1/2) ln 2

10

Show that $$\int_{a}^{b} f(x) \, dx + \int_{b}^{a} f(x) \, dx = 0$$.

[2 marks]

Solution

Using the property of swapping limits:
$$\int_{b}^{a} f(x) \, dx = -\int_{a}^{b} f(x) \, dx$$

Let $$I = \int_{a}^{b} f(x) \, dx$$. Then:
$$\int_{b}^{a} f(x) \, dx = -I$$

Therefore:
$$I + (-I) = \boxed{0}$$

Hence proved.

∴ Proved (equals 0)

Long Questions (4–5 Marks Each)

1

Evaluate $$\int_{0}^{1} x e^{x} \, dx$$ using integration by parts.

[4 marks]

Solution

Step 1 — ILATE Rule: Choose u = x (algebraic, first priority) and dv = eˣ dx (exponential, second priority).

Then du = dx and v = eˣ.

Step 2 — Apply Integration by Parts formula:
$$\int x e^{x} \, dx = x e^{x} - \int e^{x} \, dx = x e^{x} - e^{x}$$

Step 3 — Apply limits:
$$[x e^{x} - e^{x}]_{0}^{1} = (1 \cdot e^{1} - e^{1}) - (0 \cdot e^{0} - e^{0})$$
$$= (e - e) - (0 - 1) = 0 + 1 = \boxed{1}$$

∴ 1

2

Evaluate $$\int_{0}^{\pi/2} \sin^{3} x \, dx$$.

[4 marks]

Solution

Step 1: Rewrite sin³x using identity:
$$\sin^{3} x = \sin^{2} x \cdot \sin x = (1 - \cos^{2} x) \sin x$$

$$\int_{0}^{\pi/2} (1 - \cos^{2} x) \sin x \, dx$$

Step 2 — Substitution: Let u = cos x, then du = −sin x dx.
When x = 0: u = 1. When x = π/2: u = 0.

$$\int_{1}^{0} (1 - u^{2})(-du) = \int_{0}^{1} (1 - u^{2}) \, du$$

Step 3 — Integrate:
$$\left[u - \frac{u^{3}}{3}\right]_{0}^{1} = \left(1 - \frac{1}{3}\right) - 0 = \boxed{\frac{2}{3}}$$

∴ 2/3

3

Evaluate $$\int_{-1}^{2} |x^{2}-1| \, dx$$.

[5 marks]

Solution

Step 1 — Find critical points:
x² − 1 = 0 → x = ±1.
In our interval [−1, 2], the critical point is x = 1.

Step 2 — Determine sign in each sub-interval:
On (−1, 1): x² − 1 < 0, so |x²−1| = 1 − x²
On (1, 2): x² − 1 > 0, so |x²−1| = x² − 1

Step 3 — Split and evaluate:
$$\int_{-1}^{1} (1-x^{2}) \, dx + \int_{1}^{2} (x^{2}-1) \, dx$$

First integral: $$\left[x - \frac{x^{3}}{3}\right]_{-1}^{1} = \left(1 - \frac{1}{3}\right) - \left(-1 + \frac{1}{3}\right) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}$$

Second integral: $$\left[\frac{x^{3}}{3} - x\right]_{1}^{2} = \left(\frac{8}{3} - 2\right) - \left(\frac{1}{3} - 1\right) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}$$

Total: $$\frac{4}{3} + \frac{4}{3} = \boxed{\frac{8}{3}}$$

∴ 8/3

4

Evaluate $$\int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx$$ using properties.

[5 marks]

Solution

Step 1: Let $$I = \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx$$

Step 2 — Use property: $$\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$$

Here a = π/2, so f(a−x) = f(π/2 − x):
$$I = \int_{0}^{\pi/2} \frac{\sin(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)} \, dx = \int_{0}^{\pi/2} \frac{\cos x}{\cos x + \sin x} \, dx$$

Step 3 — Add the two expressions for I:
$$2I = \int_{0}^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x} \, dx = \int_{0}^{\pi/2} 1 \, dx = [x]_{0}^{\pi/2} = \frac{\pi}{2}$$

Step 4: $$I = \boxed{\frac{\pi}{4}}$$

∴ π/4

5

Evaluate $$\int_{1}^{e} x \ln x \, dx$$.

[4 marks]

Solution

Step 1 — Integration by Parts (ILATE):
u = ln x (logarithmic) → du = (1/x) dx
v = x (algebraic) → take as second function, so we integrate x to get x²/2.

$$\int x \ln x \, dx = \ln x \cdot \frac{x^{2}}{2} - \int \frac{1}{x} \cdot \frac{x^{2}}{2} \, dx$$

$$= \frac{x^{2}}{2} \ln x - \int \frac{x}{2} \, dx = \frac{x^{2}}{2} \ln x - \frac{x^{2}}{4}$$

Step 2 — Apply limits:
$$\left[\frac{x^{2}}{2} \ln x - \frac{x^{2}}{4}\right]_{1}^{e}$$

At x = e: $$\frac{e^{2}}{2} \ln e - \frac{e^{2}}{4} = \frac{e^{2}}{2} - \frac{e^{2}}{4} = \frac{e^{2}}{4}$$

At x = 1: $$\frac{1}{2} \ln 1 - \frac{1}{4} = 0 - \frac{1}{4} = -\frac{1}{4}$$

Result: $$\frac{e^{2}}{4} - \left(-\frac{1}{4}\right) = \boxed{\frac{e^{2}+1}{4}}$$

∴ (e² + 1)/4

6

Evaluate $$\int_{0}^{\pi/2} \frac{dx}{1+\sin x}$$.

[5 marks]

Solution

Step 1 — Use property: $$\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$$

Let $$I = \int_{0}^{\pi/2} \frac{dx}{1+\sin x} = \int_{0}^{\pi/2} \frac{dx}{1+\sin(\pi/2 - x)} = \int_{0}^{\pi/2} \frac{dx}{1+\cos x}$$

Step 2 — Simplify using 1 + cos x = 2cos²(x/2):
$$I = \int_{0}^{\pi/2} \frac{dx}{2\cos^{2}(x/2)} = \frac{1}{2} \int_{0}^{\pi/2} \sec^{2}\frac{x}{2} \, dx$$

Step 3 — Integrate:
Antiderivative of sec²(x/2) is 2tan(x/2).

$$\frac{1}{2} \cdot \left[2\tan\frac{x}{2}\right]_{0}^{\pi/2} = \left[\tan\frac{x}{2}\right]_{0}^{\pi/2}$$

$$= \tan\frac{\pi}{4} - \tan 0 = 1 - 0 = \boxed{1}$$

∴ 1

7

Evaluate $$\int_{0}^{a} \sqrt{a^{2}-x^{2}} \, dx$$.

[5 marks]

Solution

Step 1 — Trigonometric Substitution:
Let x = a sin θ, then dx = a cos θ dθ.

Change limits:
When x = 0: a sin θ = 0 → θ = 0
When x = a: a sin θ = a → sin θ = 1 → θ = π/2

Step 2 — Substitute:
$$\int_{0}^{\pi/2} \sqrt{a^{2} - a^{2}\sin^{2}\theta} \cdot a\cos\theta \, d\theta$$

$$= \int_{0}^{\pi/2} a\cos\theta \cdot a\cos\theta \, d\theta = a^{2} \int_{0}^{\pi/2} \cos^{2}\theta \, d\theta$$

Step 3 — Use cos²θ = (1+cos 2θ)/2:
$$a^{2} \int_{0}^{\pi/2} \frac{1 + \cos 2\theta}{2} \, d\theta = \frac{a^{2}}{2}\left[\theta + \frac{\sin 2\theta}{2}\right]_{0}^{\pi/2}$$

$$= \frac{a^{2}}{2}\left[\left(\frac{\pi}{2} + 0\right) - (0 + 0)\right] = \boxed{\frac{\pi a^{2}}{4}}$$

∴ πa²/4

8

Evaluate $$\int_{0}^{\pi} x \sin^{2} x \, dx$$.

[5 marks]

Solution

Step 1: Let $$I = \int_{0}^{\pi} x \sin^{2} x \, dx$$

Step 2 — Use property ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a−x)dx:
$$I = \int_{0}^{\pi} (\pi - x) \sin^{2}(\pi - x) \, dx = \int_{0}^{\pi} (\pi - x) \sin^{2} x \, dx$$

(since sin(π − x) = sin x)

Step 3 — Add the two expressions:
$$2I = \int_{0}^{\pi} [x \sin^{2} x + (\pi - x)\sin^{2} x] \, dx = \pi \int_{0}^{\pi} \sin^{2} x \, dx$$

Step 4 — Evaluate using sin²x = (1−cos 2x)/2:
$$2I = \pi \int_{0}^{\pi} \frac{1 - \cos 2x}{2} \, dx = \frac{\pi}{2}\left[x - \frac{\sin 2x}{2}\right]_{0}^{\pi}$$

$$= \frac{\pi}{2}\left[\left(\pi - 0\right) - (0 - 0)\right] = \frac{\pi^{2}}{2}$$

Step 5: $$I = \boxed{\frac{\pi^{2}}{4}}$$

∴ π²/4

9

Evaluate $$\int_{0}^{2} \frac{x}{\sqrt{x^{2}+5}} \, dx$$.

[4 marks]

Solution

Step 1 — Substitution: Let u = x² + 5, then du = 2x dx, so x dx = du/2.

Change limits:
When x = 0: u = 0 + 5 = 5
When x = 2: u = 4 + 5 = 9

Step 2 — Rewrite and integrate:
$$\int_{5}^{9} \frac{1}{2\sqrt{u}} \, du = \frac{1}{2} \int_{5}^{9} u^{-1/2} \, du$$

$$= \frac{1}{2} \left[\frac{u^{1/2}}{1/2}\right]_{5}^{9} = \frac{1}{2} \cdot 2[\sqrt{u}]_{5}^{9} = [\sqrt{u}]_{5}^{9}$$

$$= \sqrt{9} - \sqrt{5} = \boxed{3 - \sqrt{5}}$$

∴ 3 − √5

10

Evaluate $$\int_{0}^{2\pi} |\cos x| \, dx$$.

[5 marks]

Solution

Step 1 — Find where cos x = 0:
In [0, 2π], cos x = 0 at x = π/2 and x = 3π/2.

Step 2 — Determine sign in each interval:
(0, π/2): cos x > 0
(π/2, 3π/2): cos x < 0
(3π/2, 2π): cos x > 0

Step 3 — Split the integral:
$$\int_{0}^{\pi/2} \cos x \, dx - \int_{\pi/2}^{3\pi/2} \cos x \, dx + \int_{3\pi/2}^{2\pi} \cos x \, dx$$

Step 4 — Evaluate each part:
$$[\sin x]_{0}^{\pi/2} = \sin\frac{\pi}{2} - \sin 0 = 1 - 0 = 1$$

$$-[\sin x]_{\pi/2}^{3\pi/2} = -(\sin\frac{3\pi}{2} - \sin\frac{\pi}{2}) = -(-1 - 1) = 2$$

$$[\sin x]_{3\pi/2}^{2\pi} = \sin 2\pi - \sin\frac{3\pi}{2} = 0 - (-1) = 1$$

Total: $$1 + 2 + 1 = \boxed{4}$$

∴ 4

Chapter Test — Full Mixed Paper

Attempt this test in 50 minutes without notes. This mirrors the NEB exam format for Definite Integrals.

1 Section A — MCQs (1 Mark Each)

Mixed Level — MCQ

1

$$\int_{0}^{1} (3x^{2}+1) \, dx$$ equals?

Correct!

Incorrect. Correct: 2

Solution

$$\int_{0}^{1} (3x^{2}+1) \, dx = [x^{3} + x]_{0}^{1} = (1+1) - (0+0) = \boxed{2}$$

∴ 2

2

Is sin x even or odd?

Correct!

Incorrect. Correct: Odd

Solution

Check: sin(−x) = −sin(x).

Since f(−x) = −f(x), sin x is an odd function.

∴ Odd

3

$$\int_{1}^{2} \frac{1}{x} \, dx = ?$$

Correct!

Incorrect. Correct: ln 2

Solution

$$\int_{1}^{2} \frac{1}{x} \, dx = [\ln|x|]_{1}^{2} = \ln 2 - \ln 1 = \ln 2 - 0 = \boxed{\ln 2}$$

∴ ln 2

2 Section B — Short Questions (2 Marks Each)

1

Evaluate $$\int_{0}^{\pi/4} \sec^{2} x \, dx$$.

[2 marks]

Solution

Antiderivative of sec²x is tan x.

$$[\tan x]_{0}^{\pi/4} = \tan\frac{\pi}{4} - \tan 0 = 1 - 0 = \boxed{1}$$

∴ 1

2

Evaluate $$\int_{-1}^{1} |x| \, dx$$.

[2 marks]

Solution

Split at x = 0 where |x| changes sign:

$$\int_{-1}^{0} (-x) \, dx + \int_{0}^{1} x \, dx = \left[-\frac{x^{2}}{2}\right]_{-1}^{0} + \left[\frac{x^{2}}{2}\right]_{0}^{1}$$

$$= (0 + \frac{1}{2}) + (\frac{1}{2} - 0) = \boxed{1}$$

Alternatively: |x| is even, so 2∫₀¹ x dx = 2 × (1/2) = 1.

∴ 1

3

Evaluate $$\int_{1}^{2} (x^{2}-3) \, dx$$.

[2 marks]

Solution

$$\left[\frac{x^{3}}{3} - 3x\right]_{1}^{2} = \left(\frac{8}{3} - 6\right) - \left(\frac{1}{3} - 3\right)$$

$$= \frac{8}{3} - 6 - \frac{1}{3} + 3 = \frac{7}{3} - 3 = \frac{7-9}{3} = \boxed{-\frac{2}{3}}$$

∴ −2/3

3 Section C — Long Questions (4–5 Marks Each)

1

Evaluate $$\int_{0}^{\pi/2} \sin^{2} x \cos^{2} x \, dx$$.

[5 marks]

Solution

Step 1: Use identity $$\sin^{2} x \cos^{2} x = \frac{(2\sin x \cos x)^{2}}{4} = \frac{\sin^{2} 2x}{4}$$

Step 2: Use $$\sin^{2} 2x = \frac{1 - \cos 4x}{2}$$:
$$\frac{1}{4} \int_{0}^{\pi/2} \frac{1 - \cos 4x}{2} \, dx = \frac{1}{8} \int_{0}^{\pi/2} (1 - \cos 4x) \, dx$$

Step 3: $$\frac{1}{8}\left[x - \frac{\sin 4x}{4}\right]_{0}^{\pi/2} = \frac{1}{8}\left[\left(\frac{\pi}{2} - \frac{\sin 2\pi}{4}\right) - (0 - 0)\right]$$

$$= \frac{1}{8} \times \frac{\pi}{2} = \boxed{\frac{\pi}{16}}$$

∴ π/16

2

Evaluate $$\int_{0}^{1} \frac{\tan^{-1} x}{1+x^{2}} \, dx$$.

[4 marks]

Solution

Step 1 — Substitution: Let u = tan⁻¹x, then $$du = \frac{1}{1+x^{2}} dx$$.

Change limits:
When x = 0: u = tan⁻¹0 = 0
When x = 1: u = tan⁻¹1 = π/4

Step 2: $$\int_{0}^{\pi/4} u \, du = \left[\frac{u^{2}}{2}\right]_{0}^{\pi/4} = \frac{1}{2}\left(\frac{\pi^{2}}{16} - 0\right) = \boxed{\frac{\pi^{2}}{32}}$$

∴ π²/32

3

Prove that $$\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$$.

[4 marks]

Solution

Step 1: Let $$I = \int_{0}^{a} f(x) \, dx$$

Step 2 — Substitution: Let t = a − x, then dt = −dx.

When x = 0: t = a
When x = a: t = 0

Step 3: $$I = \int_{a}^{0} f(a-t)(-dt) = \int_{0}^{a} f(a-t) \, dt$$

Since t is a dummy variable, replace t with x:
$$\int_{0}^{a} f(a-x) \, dx = \boxed{\text{RHS = LHS}}$$

Hence proved.

∴ Proved

One-Page Revision Cheat Sheet

1

Key Formula: $$\int_{a}^{b} f(x) \, dx = F(b) - F(a)$$ — always upper minus lower.

2

Never add +c: Definite integrals give a number. The constant cancels out.

3

Symmetry Shortcut: Even function → $$2\int_{0}^{a}$$; Odd function → 0 (for limits −a to a).

4

Modulus Rule: Always split the integral where the inside expression equals 0.

5

Substitution Rule: Always change limits when substituting u = g(x).

6

Property of Identical Limits: $$\int_{a}^{a} f(x) \, dx = 0$$.

7

Swapping Limits: $$\int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx$$.

8

Interval Splitting: $$\int_{a}^{b} = \int_{a}^{c} + \int_{c}^{b}$$ for any c between a and b.

9

Property: $$\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$$ — essential for trig fractions.

10

Common Mistake: Forgetting e⁰ = 1 during evaluation — not 0!

11

Last Minute Tip: If a trig integral looks impossible, try the f(a−x) property immediately!

12

Quick Check: Is your final answer a number? If yes, you're on the right track. If it still has x or +c, recheck.