Definite integrals questions with solutions are given here for practice, solving these questions will be helpful for understanding various properties of definite integrals. A definite integral is of the form,

\begin{array}{l} \int_{a}^{b} f (x) d x = F (b) - F (a) \end{array}

Where the function f is a continuous function within an interval [a, b] and F is the antiderivative of f. In other words, the definite integral of a function f means

Integration of f between a to b = value of the antiderivative of f at b (upper limit) – value of the antiderivative of f at a (lower limit).

It is represented graphically as

Definite integration

Thus, integrating the function f from a to b means finding the area under the curve f(x) from x = a to x = b.

Learn more about Definite Integrals.

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Definite Integrals Questions with Solutions

Let us solve a few definite integral questions given below. Also, check your answers with the solutions provided.

Question 1:

Evaluate the following integral:

\begin{array}{l} \int_{0}^{π / 2} c o s^{4} x d x \end{array}

Solution:

\begin{array}{l} (i) \int_{0}^{π / 2} c o s^{4} x d x = \int_{0}^{π / 2} (c o s^{2} x)^{2} d x \end{array}

\begin{array}{l} = \int_{0}^{π / 2} {(\frac{1 + c o s 2 x}{2})}^{2} d x = \frac{1}{4} \int_{0}^{π / 2} (1 + 2 c o s 2 x + c o s^{2} 2 x) d x \end{array}

\begin{array}{l} = \frac{1}{4} \int_{0}^{π / 2} 1. d x + \frac{2}{4} \int_{0}^{π / 2} c o s 2 x d x + \frac{1}{4} \int_{0}^{π / 2} (\frac{1 + c o s 4 x}{2}) d x \end{array}

\begin{array}{l} = \frac{π}{8} + \frac{1}{2} {[\frac{s i n 2 x}{2}]}_{0}^{π / 2} + \frac{1}{8} \int_{0}^{π / 2} 1. d x + \frac{1}{8} \int_{0}^{π / 2} c o s 4 x d x \end{array}

\begin{array}{l} = \frac{π}{8} + \frac{1}{2} [0 - 0] + {[\frac{x}{8}]}_{0}^{π / 2} + \frac{1}{32} [s i n 4 x]_{0}^{π / 2} = \frac{π}{8} + \frac{π}{16} + \frac{1}{32} [0 - 0] \end{array}

\begin{array}{l} ∴ \int_{0}^{π / 2} c o s^{4} x d x = \frac{3 π}{16} \end{array}

Question 2:

Evaluate the following integral:

\begin{array}{l} \int_{0}^{π / 2} \sqrt{1 + s i n 2 x} d x \end{array}

Solution:

\begin{array}{l} \int_{0}^{π / 2} \sqrt{1 + s i n 2 x} d x = \int_{0}^{π / 2} \sqrt{1 + c o s (\frac{π}{2} - 2 x)} d x \end{array}

\begin{array}{l} = \int_{0}^{π / 2} \sqrt{2 c o s^{2} (\frac{π}{4} - x)} d x = \sqrt{2} \int_{0}^{π / 2} c o s (\frac{π}{4} - x) d x \end{array}

Let (𝜋/4 – x) = t, dx = – dt. Also, when x = 0 ⇒ t = 𝜋/4 and x = 𝜋/2 ⇒ t = – 𝜋/4

\begin{array}{l} = - \sqrt{2} \int_{π / 4}^{- π / 4} c o s t d x = 2 \sqrt{2} \int_{- π / 4}^{π / 4} c o s t d x = 2 \sqrt{2} \int_{0}^{π / 4} c o s t d x \end{array}

\begin{array}{l} = 2 \sqrt{2} {[s i n t]}_{0}^{π / 4} = 2 \sqrt{2} . \frac{1}{\sqrt{2}} \end{array}

\begin{array}{l} ∴ \int_{0}^{π / 2} \sqrt{1 + s i n 2 x} d x = 2 \end{array}

Question 3:

Evaluate the following integral:

\begin{array}{l} \int_{0}^{π / 4} {(t a n x + c o t x)}^{- 2} d x \end{array}

Solution:

\begin{array}{l} \int_{0}^{π / 4} {(t a n x + c o t x)}^{- 2} d x = \int_{0}^{π / 4} {(\frac{s i n x}{c o s x} + \frac{c o s x}{s i n x})}^{- 2} d x = \int_{0}^{π / 4} {(\frac{s i n^{2} x + c o s^{2} x}{s i n x c o s x})}^{- 2} d x \end{array}

\begin{array}{l} = \frac{1}{4} \int_{0}^{π / 4} {(2 s i n x c o s x)}^{2} d x = \frac{1}{4} \int_{0}^{π / 4} s i n^{2} 2 x d x = \frac{1}{4} \int_{0}^{π / 4} \frac{1 - c o s 4 x}{2} d x \end{array}

\begin{array}{l} = \frac{1}{8} {[x]}_{0}^{π / 4} + \frac{1}{8} {[\frac{s i n 4 x}{4}]}_{0}^{π / 4} \end{array}

\begin{array}{l} ∴ \int_{0}^{π / 4} {(t a n x + c o t x)}^{- 2} d x = \frac{π}{32} \end{array}

Question 4:

Evaluate the following integral:

\begin{array}{l} \int_{0}^{π / 2} x . s i n^{2} x d x \end{array}

Solution:

\begin{array}{l} \int_{0}^{π / 2} x . s i n^{2} x d x = \int_{0}^{π / 2} x . (\frac{1 - c o s 2 x}{2}) d x \end{array}

\begin{array}{l} = \int_{0}^{π / 2} \frac{x}{2} d x - \frac{1}{2} \int_{0}^{π / 2} x . c o s 2 x d x = {[\frac{x^{2}}{4}]}_{0}^{π / 2} - [{(\frac{x . s i n 2 x}{2})}_{0}^{π / 2} + \frac{1}{4} {(c o s 2 x)}_{0}^{π / 2}] \end{array}

\begin{array}{l} = \frac{π^{2}}{16} + \frac{1}{4} \end{array}

\begin{array}{l} ∴ \int_{0}^{π / 2} x . s i n^{2} x d x = \frac{π^{2}}{16} + \frac{1}{4} \end{array}

Question 5:

Evaluate the following integral:

\begin{array}{l} \int_{π / 4}^{π / 2} c o s 2 x . l o g (s i n x) d x \end{array}

Solution:

Given,

\begin{array}{l} \int_{π / 4}^{π / 2} c o s 2 x . l o g (s i n x) d x \end{array}

Using integration by parts, we have,

\begin{array}{l} \int_{π / 4}^{π / 2} c o s 2 x . l o g (s i n x) d x = {[l o g (s i n x) \frac{s i n 2 x}{2}]}_{π / 4}^{π / 2} - \int_{π / 4}^{π / 2} \frac{c o s x}{s i n x} . \frac{s i n 2 x}{2} d x \end{array}

\begin{array}{l} = l o g 1 \frac{s i n π}{2} - \frac{1}{2} l o g \frac{1}{\sqrt{2}} - \int_{π / 4}^{π / 2} (\frac{1 + c o s 2 x}{2}) d x \end{array}

\begin{array}{l} = 0 + \frac{1}{4} l o g 2 - {[\frac{x}{2} + \frac{s i n 2 x}{4}]}_{π / 4}^{π / 2} \end{array}

\begin{array}{l} ∴ \int_{π / 4}^{π / 2} c o s 2 x . l o g (s i n x) d x = \frac{1}{4} l o g 2 - \frac{π}{8} + \frac{1}{4} \end{array}

Also Read:

Question 6:

Evaluate the following integral:

\begin{array}{l} \int_{0}^{1} \frac{x . e^{x}}{(x + 1)^{2}} d x \end{array}

Solution:

We have,

\begin{array}{l} \int_{0}^{1} \frac{x . e^{x}}{(x + 1)^{2}} d x = \int_{0}^{1} e^{x} \frac{(x + 1) - 1}{(x + 1)^{2}} d x \end{array}

\begin{array}{l} = \int_{0}^{1} \frac{e^{x}}{(x + 1)} d x - \int_{0}^{1} \frac{e^{x}}{(x + 1)^{2}} d x \end{array}

Integrating the first integral by parts taking (x + 1)^–1 as the first function,

\begin{array}{l} = {[\frac{e^{x}}{x + 1}]}_{0}^{1} - \int_{0}^{1} (- 1) . (x + 1)^{- 2} . e^{x} d x - \int_{0}^{1} \frac{e^{x}}{(x + 1)^{2}} d x \end{array}

\begin{array}{l} = [\frac{e}{2} - 1] + \int_{0}^{1} \frac{e^{x}}{(x + 1)^{2}} d x - \int_{0}^{1} \frac{e^{x}}{(x + 1)^{2}} d x \end{array}

\begin{array}{l} ∴ \int_{0}^{1} \frac{x . e^{x}}{(x + 1)^{2}} d x = [\frac{e}{2} - 1] \end{array}

Question 7:

Evaluate the following integral:

\begin{array}{l} \int_{0}^{π / 2} \sqrt{t a n x} + \sqrt{c o t x} d x \end{array}

Solution:

We have,

\begin{array}{l} \int_{0}^{π / 2} \sqrt{t a n x} + \sqrt{c o t x} d x = \int_{0}^{π / 2} \sqrt{\frac{s i n x}{c o s x}} + \sqrt{\frac{c o s x}{s i n x}} d x \end{array}

\begin{array}{l} = \int_{0}^{π / 2} \frac{s i n x + c o s x}{\sqrt{s i n x c o s x}} d x = \sqrt{2} \int_{0}^{π / 2} \frac{s i n x + c o s x}{\sqrt{2 s i n x c o s x}} d x = \sqrt{2} \int_{0}^{π / 2} \frac{s i n x + c o s x}{\sqrt{s i n 2 x}} d x \end{array}

Let sin x – cos x = t, (cos x + sin x) dx = dt

Again, sin²x + cos²x – 2sin x cos x = t² ⇒ 1 – sin 2x = t² ⇒ sin 2x = 1 – t²

Now, when x = 0 ⇒ t = –1

And x = 𝜋/2 ⇒ t = 1

\begin{array}{l} = \sqrt{2} \int_{- 1}^{1} \frac{d t}{\sqrt{1 - t^{2}}} = \sqrt{2} {[s i n^{- 1} t]}_{- 1}^{1} \end{array}

\begin{array}{l} ∴ \int_{0}^{π / 2} \sqrt{t a n x} + \sqrt{c o t x} d x = \sqrt{2} π \end{array}

Question 8:

Evaluate using the properties of definite integral:

\begin{array}{l} \int_{0}^{π / 2} \frac{d x}{1 + \sqrt{t a n x}} \end{array}

Solution:

Let,

\begin{array}{l} I = \int_{0}^{π / 2} \frac{d x}{1 + \sqrt{t a n x}} = \int_{0}^{π / 2} \frac{\sqrt{c o s x} d x}{\sqrt{c o s x} + \sqrt{s i n x}} \dots . (i) \end{array}

Now, using the property,

\begin{array}{l} \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x \end{array}

Then,

\begin{array}{l} I = \int_{0}^{π / 2} \frac{\sqrt{c o s (\frac{π}{2} - x)} d x}{\sqrt{c o s (\frac{π}{2} - x)} + \sqrt{s i n (\frac{π}{2} - x)}} = \int_{0}^{π / 2} \frac{\sqrt{s i n x} d x}{\sqrt{s i n x} + \sqrt{s i n x}} \dots (i i) \end{array}

Adding (i) and (ii), we get,

\begin{array}{l} 2 I = \int_{0}^{π / 2} \frac{\sqrt{c o s x} + \sqrt{s i n x} d x}{\sqrt{s i n x} + \sqrt{s i n x}} = \int_{0}^{π / 2} 1. d x = π / 2 \end{array}

Or, I = 𝜋/4

\begin{array}{l} ∴ \int_{0}^{π / 2} \frac{d x}{1 + \sqrt{t a n x}} = \frac{π}{4} \end{array}

Also Refer:

Question 9:

\begin{array}{l} \int_{0}^{a} \sqrt{x} d x = 2 a \int_{0}^{π / 2} s i n^{3} x d x \end{array}

find the value of

\begin{array}{l} \int_{a}^{a + 1} x^{2} d x \end{array}

Solution:

Now, given

\begin{array}{l} \int_{0}^{a} \sqrt{x} d x = 2 a \int_{0}^{π / 2} s i n^{3} x d x \end{array}

\begin{array}{l} \Rightarrow {[\frac{x^{3 / 2}}{3 / 2}]}_{0}^{a} = 2 a \int_{0}^{π / 2} [\frac{3 s i n x - s i n 3 x}{4}] d x (∵ s i n 3 x = 3 s i n x - 4 s i n^{3} x) \end{array}

\begin{array}{l} \Rightarrow \frac{2}{3} a^{3 / 2} = 2 a {[\frac{- 3 c o s x}{4} + \frac{c o s 3 x}{12}]}_{0}^{π / 2} \end{array}

\begin{array}{l} \Rightarrow \frac{2}{3} . \frac{1}{2} a^{3 / 2 - 1} = \frac{3}{4} - \frac{1}{12} \end{array}

\begin{array}{l} \Rightarrow \sqrt{a} = 2 \Rightarrow a = 4 \end{array}

Then,

\begin{array}{l} \int_{a}^{a + 1} x^{2} d x = \int_{4}^{5} x^{2} d x = {[\frac{x^{2}}{3}]}_{4}^{5} \end{array}

\begin{array}{l} = [\frac{125}{3} - \frac{64}{3}] = \frac{61}{3} \end{array}

Question 10:

Let f: R → R be a differentiable function, if f(1) = 4 then prove that

\begin{array}{l} lim_{x \to 1} \int_{4}^{f (x)} \frac{2 t}{x - 1} d t = 8 f^{'} (1) \end{array}

Solution:

Now,

\begin{array}{l} lim_{x \to 1} \int_{4}^{f (x)} \frac{2 t}{x - 1} d t = lim_{x \to 1} \frac{1}{x - 1} \int_{4}^{f (x)} 2 t d t = lim_{x \to 1} \frac{1}{x - 1} {[t^{2}]}_{4}^{f (x)} \end{array}

\begin{array}{l} = lim_{x \to 1} \frac{{f (x)}^{2} - 4^{2}}{x - 1} = lim_{x \to 1} [f (x) + 4] . lim_{x \to 1} \frac{f (x) - 4}{x - 1} \end{array}

\begin{array}{l} = [f (1) + 4] . lim_{x \to 1} \frac{f (x) - f (1)}{x - 1} [∵ f (1) = 4] \end{array}

\begin{array}{l} = 8. lim_{x \to 1} \frac{f (x) - f (1)}{x - 1} \end{array}

Let x = 1 + h, where is a very small positive quantity such that when h → 0 ⇒ x → 1, then we have,

\begin{array}{l} = 8. lim_{h \to 0} \frac{f (1 + h) - f (1)}{h} \end{array}

By the definition of derivative of a function at any point, we get

\begin{array}{l} lim_{x \to 1} \int_{4}^{f (x)} \frac{2 t}{x - 1} d t = 8 f^{'} (1) . \end{array}

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Practice Questions on Definite Integrals

1. Evaluate the following integrals:

\begin{array}{l} (i) \int_{0}^{π / 4} \sqrt{1 - s i n 2 x} d x \end{array}

\begin{array}{l} (i i) \int_{0}^{1} x^{2} e^{x} d x \end{array}

\begin{array}{l} (i i i) \int_{0}^{1} x . t a n^{- 1} x d x \end{array}

\begin{array}{l} (i v) \int_{1}^{5} \frac{l o g x}{(x + 1)^{2}} d x \end{array}

2. Evaluate using the properties of definite integral:

\begin{array}{l} (i) \int_{0}^{2} x^{2} \sqrt{2 - x} d x \end{array}

\begin{array}{l} (i i) \int_{0}^{2 π} \frac{x . s i n^{2 n} x d x}{s i n^{2 n} x + c o s^{2 n} x}, n ϵ N \end{array}

3. If

\begin{array}{l} \int_{a}^{b} x^{3} d x = 0 a n d \int_{a}^{b} x^{2} d x = \frac{2}{3}, \end{array}

find the value of a and b.

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