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,
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\(\begin{array}{l}\int_{a}^{b}f(x)dx=F(b)-F(a)\end{array} \)
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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
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.
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:
\(\begin{array}{l}\int_{0}^{\pi/2}cos^{4}x\:dx\end{array} \)
Solution:
\(\begin{array}{l}(i)\:\int_{0}^{\pi/2}cos^{4}x\:dx=\int_{0}^{\pi/2}(cos^{2}x)^{2}\:dx\end{array} \)
\(\begin{array}{l}=\int_{0}^{\pi /2}\left ( \frac{1+cos\:2x}{2} \right )^{2}\:dx=\frac{1}{4}\int_{0}^{\pi /2}\left ( 1+2cos\:2x+cos^{2}2x \right )dx\end{array} \)
\(\begin{array}{l}=\frac{1}{4}\int_{0}^{\pi/2}1.dx+\frac{2}{4}\int_{0}^{\pi/2}cos\:2x\:dx+\frac{1}{4}\int_{0}^{\pi/2}\left ( \frac{1+cos\:4x}{2} \right )dx\end{array} \)
\(\begin{array}{l}=\frac{\pi}{8}+\frac{1}{2}\left [ \frac{sin\:2x}{2} \right ]_{0}^{\pi/2}+\frac{1}{8}\int _{0}^{\pi/2}1.dx+\frac{1}{8}\int_{0}^{\pi/2}cos\:4x\:dx\end{array} \)
\(\begin{array}{l}=\frac{\pi}{8}+\frac{1}{2}[0-0]+\left [ \frac{x}{8} \right ]_{0}^{\pi/2}+\frac{1}{32}[sin\:4x]_{0}^{\pi/2}=\frac{\pi}{8}+\frac{\pi}{16}+\frac{1}{32}[0-0]\end{array} \)
\(\begin{array}{l}\therefore \int_{0}^{\pi/2}cos^{4}x\:dx=\frac{3\pi}{16}\end{array} \)
Question 2:
Evaluate the following integral:
\(\begin{array}{l}\int_{0}^{\pi/2}\sqrt{1+sin\:2x}\:dx\end{array} \)
Solution:
\(\begin{array}{l}\int_{0}^{\pi/2}\sqrt{1+sin\:2x}\:dx=\int_{0}^{\pi/2}\sqrt{1+cos\left ( \frac{\pi}{2}-2x \right )}\:dx\end{array} \)
\(\begin{array}{l}=\int_{0}^{\pi/2}\sqrt{2cos^{2}\left ( \frac{\pi}{4}-x \right )}\:dx=\sqrt{2}\int_{0}^{\pi/2}cos\left ( \frac{\pi}{4}-x \right )\:dx\end{array} \)
Let (𝜋/4 – x) = t, dx = – dt. Also, when x = 0 ⇒ t = 𝜋/4 and x = 𝜋/2 ⇒ t = – 𝜋/4
\(\begin{array}{l}=-\sqrt{2}\int_{\pi/4}^{-\pi/4}cos\:t\:dx=2\sqrt{2}\int_{-\pi/4}^{\pi/4}cos\:t\:dx=2\sqrt{2}\int_{0}^{\pi/4}cos\:t\:dx\end{array} \)
\(\begin{array}{l}=2\sqrt{2}\left [ sin\:t \right ]_{0}^{\pi/4}=2\sqrt{2}.\frac{1}{\sqrt{2}}\end{array} \)
\(\begin{array}{l}\therefore \int_{0}^{\pi/2}\sqrt{1+sin\:2x}\:dx=2\end{array} \)
Question 3:
Evaluate the following integral:
\(\begin{array}{l}\int_{0}^{\pi/4}\left ( tan\:x+cot\:x \right )^{-2}\:dx\end{array} \)
Solution:
\(\begin{array}{l} \int_{0}^{\pi/4}\left ( tan\:x+cot\:x \right )^{-2}\:dx= \int_{0}^{\pi/4}\left ( \frac{sin\:x}{cos\:x}+\frac{cos\:x}{sin\:x} \right )^{-2}\:dx=\int_{0}^{\pi/4}\left ( \frac{sin^{2}x+cos^{2}x}{sin\:x cos\:x} \right )^{-2}\:dx\end{array} \)
\(\begin{array}{l}=\frac{1}{4}\int_{0}^{\pi/4}\left (2\:sin\:x \:cos\:x \right )^{2}\:dx=\frac{1}{4}\int_{0}^{\pi/4}sin^{2}\:2x\:dx=\frac{1}{4}\int_{0}^{\pi/4}\frac{1-cos\:4x}{2}\:dx\end{array} \)
\(\begin{array}{l}=\frac{1}{8}\left [ x \right ]_{0}^{\pi/4}+\frac{1}{8}\left [ \frac{sin\:4x}{4} \right ]_{0}^{\pi/4}\end{array} \)
\(\begin{array}{l}\therefore \int_{0}^{\pi/4}\left ( tan\:x+cot\:x \right )^{-2}\:dx=\frac{\pi}{32}\end{array} \)
Question 4:
Evaluate the following integral:
\(\begin{array}{l}\int_{0}^{\pi/2}x.sin^{2}x\:dx\end{array} \)
Solution:
\(\begin{array}{l}\int_{0}^{\pi/2}x.sin^{2}x\:dx=\int_{0}^{\pi/2}x.\left ( \frac{1-cos\:2x}{2} \right )\:dx\end{array} \)
\(\begin{array}{l}=\int_{0}^{\pi/2}\frac{x}{2}\:dx-\frac{1}{2}\int_{0}^{\pi/2}x.cos\:2x\:dx=\left [ \frac{x^{2}}{4} \right ]_{0}^{\pi/2}-\left [ \left ( \frac{x.sin\:2x}{2} \right )_{0}^{\pi/2}+\frac{1}{4}\left ( cos\:2x \right )_{0}^{\pi/2} \right ]\end{array} \)
\(\begin{array}{l}= \frac{\pi^{2}}{16} +\frac{1}{4}\end{array} \)
\(\begin{array}{l}\therefore \int_{0}^{\pi/2}x.sin^{2}x\:dx= \frac{\pi^{2}}{16} +\frac{1}{4}\end{array} \)
Question 5:
Evaluate the following integral:
\(\begin{array}{l}\int_{\pi/4}^{\pi/2}cos\:2x.log(sin\:x)\:dx\end{array} \)
Solution:
Given,
\(\begin{array}{l}\int_{\pi/4}^{\pi/2}cos\:2x.log(sin\:x)\:dx\end{array} \)
Using integration by parts, we have,
\(\begin{array}{l}\int_{\pi/4}^{\pi/2}cos\:2x.log(sin\:x)\:dx=\left [ log(sin\:x)\frac{sin\:2x}{2} \right ]_{\pi/4}^{\pi/2}-\int _{\pi/4}^{\pi/2}\frac{cos\:x}{sin\:x}.\frac{sin\:2x}{2}dx\end{array} \)
\(\begin{array}{l}=log\:1\frac{sin\:\pi}{2}-\frac{1}{2}log\frac{1}{\sqrt{2}}-\int_{\pi/4}^{\pi/2}\left ( \frac{1+cos\:2x}{2} \right )\:dx\end{array} \)
\(\begin{array}{l}=0+\frac{1}{4}log\:2-\left [ \frac{x}{2}+\frac{sin\:2x}{4} \right ]_{\pi/4}^{\pi/2}\end{array} \)
\(\begin{array}{l}\therefore \int_{\pi/4}^{\pi/2}cos\:2x.log(sin\:x)\:dx=\frac{1}{4}log\:2-\frac{\pi}{8}+\frac{1}{4}\end{array} \)
Also Read:
Question 6:
Evaluate the following integral:
\(\begin{array}{l}\int_{0}^{1}\frac{x.e^{x}}{(x+1)^{2}}dx\end{array} \)
Solution:
We have,
\(\begin{array}{l}\int_{0}^{1}\frac{x.e^{x}}{(x+1)^{2}}dx=\int_{0}^{1}e^{x}\frac{(x+1)-1}{(x+1)^{2}}dx\end{array} \)
\(\begin{array}{l}=\int_{0}^{1}\frac{e^{x}}{(x+1)}dx-\int_{0}^{1}\frac{e^{x}}{(x+1)^{2}}dx\end{array} \)
Integrating the first integral by parts taking (x + 1) –1 as the first function,
\(\begin{array}{l}=\left [ \frac{e^{x}}{x+1} \right ]_{0}^{1}-\int_{0}^{1}(-1).(x+1)^{-2}.e^{x}dx-\int_{0}^{1}\frac{e^{x}}{(x+1)^{2}}dx\end{array} \)
\(\begin{array}{l}=\left [ \frac{e}{2}-1 \right ]+\int_{0}^{1}\frac{e^{x}}{(x+1)^{2}}dx-\int_{0}^{1}\frac{e^{x}}{(x+1)^{2}}dx\end{array} \)
\(\begin{array}{l}\therefore \int_{0}^{1}\frac{x.e^{x}}{(x+1)^{2}}dx=\left [ \frac{e}{2}-1 \right ]\end{array} \)
Question 7:
Evaluate the following integral:
\(\begin{array}{l}\int_{0}^{\pi/2}\sqrt{tan\:x}+\sqrt{cot\:x}\:dx\end{array} \)
Solution:
We have,
\(\begin{array}{l}\int_{0}^{\pi/2}\sqrt{tan\:x}+\sqrt{cot\:x}\:dx=\int_{0}^{\pi/2}\sqrt{\frac{sin\:x}{cos\:x}}+\sqrt{\frac{cos\:x}{sin\:x}}\:dx\end{array} \)
\(\begin{array}{l}=\int_{0}^{\pi/2}\frac{sin\:x+cos\:x}{\sqrt{sin\:x\:cos\:x}}\:dx=\sqrt{2}\int_{0}^{\pi/2}\frac{sin\:x+cos\:x}{\sqrt{2sin\:x\:cos\:x}}\:dx=\sqrt{2}\int_{0}^{\pi/2}\frac{sin\:x+cos\:x}{\sqrt{sin\:2x}}\:dx\end{array} \)
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
\(\begin{array}{l}=\sqrt{2}\int_{-1}^{1}\frac{dt}{\sqrt{1-t^{2}}}= \sqrt{2}\left [ sin^{-1}t \right ]_{-1}^{1}\end{array} \)
\(\begin{array}{l}\therefore \int_{0}^{\pi/2}\sqrt{tan\:x}+\sqrt{cot\:x}\:dx=\sqrt{2}\pi\end{array} \)
Question 8:
Evaluate using the properties of definite integral:
\(\begin{array}{l}\int_{0}^{\pi/2}\frac{dx}{1+\sqrt{tan\:x}}\end{array} \)
Solution:
Let,
\(\begin{array}{l}I=\int_{0}^{\pi/2}\frac{dx}{1+\sqrt{tan\:x}}=\int_{0}^{\pi/2}\frac{\sqrt{cos\:x}\:dx}{\sqrt{cos\:x}+\sqrt{sin\:x}}\:\:\:\:….(i)\end{array} \)
Now, using the property,
\(\begin{array}{l}\int_{0}^{a}f(x)\:dx=\int_{0}^{a}f(a-x)\:dx\end{array} \)
Then,
\(\begin{array}{l}I=\int_{0}^{\pi/2}\frac{\sqrt{cos\:\left ( \frac{\pi}{2}-x \right )}\:dx}{\sqrt{cos\:\left ( \frac{\pi}{2}-x \right )}+\sqrt{sin\:\left ( \frac{\pi}{2}-x \right )}}=\int_{0}^{\pi/2}\frac{\sqrt{sin\:x}\:dx}{\sqrt{sin\:x}+\sqrt{sin\:x}}\:\:\:\:…(ii)\end{array} \)
Adding (i) and (ii), we get,
\(\begin{array}{l}2I=\int_{0}^{\pi/2}\frac{\sqrt{cos\:x}+\sqrt{sin\:x}\:dx}{\sqrt{sin\:x}+\sqrt{sin\:x}}=\int_{0}^{\pi/2}1.dx=\pi/2\end{array} \)
Or, I = 𝜋/4
\(\begin{array}{l}\therefore \int_{0}^{\pi/2}\frac{dx}{1+\sqrt{tan\:x}}=\frac{\pi}{4}\end{array} \)
Also Refer:
Question 9:
If
\(\begin{array}{l}\int_{0}^{a}\sqrt{x}\:dx=2a\int_{0}^{\pi/2}sin^{3}x\:dx\end{array} \)
find the value of \(\begin{array}{l}\int_{a}^{a+1}x^{2}\:dx\end{array} \)
Solution:
Now, given
\(\begin{array}{l}\int_{0}^{a}\sqrt{x}\:dx=2a\int_{0}^{\pi/2}sin^{3}x\:dx\end{array} \)
\(\begin{array}{l}\Rightarrow\left [ \frac{x^{3/2}}{3/2}\ \right ]_{0}^{a}=2a\int_{0}^{\pi/2}\left [ \frac{3sin\:x-sin\:3x}{4} \right ]\:dx\:\:\:\:\left ( \because sin\:3x=3\:sin\:x-4\:sin^{3}x \right )\end{array} \)
\(\begin{array}{l}\Rightarrow \frac{2}{3}a^{3/2}=2a\left [ \frac{-3cos\:x}{4}+\frac{cos\:3x}{12} \right ]_{0}^{\pi/2}\end{array} \)
\(\begin{array}{l}\Rightarrow \frac{2}{3}.\frac{1}{2}a^{3/2-1}=\frac{3}{4}-\frac{1}{12}\end{array} \)
\(\begin{array}{l}\Rightarrow \sqrt{a}=2\Rightarrow a=4\end{array} \)
Then,
\(\begin{array}{l}\int_{a}^{a+1}x^{2}\:dx=\int_{4}^{5}x^{2}\:dx=\left [ \frac{x^{2}}{3} \right ]_{4}^{5}\end{array} \)
\(\begin{array}{l}=\left [ \frac{125}{3}-\frac{64}{3} \right ]=\frac{61}{3}\end{array} \)
Question 10:
Let f: R → R be a differentiable function, if f(1) = 4 then prove that
\(\begin{array}{l}\displaystyle \lim_{x \to 1}\int_{4}^{f(x)}\frac{2t}{x-1}dt=8f'(1)\end{array} \)
Solution:
Now,
\(\begin{array}{l}\displaystyle \lim_{x \to 1}\int_{4}^{f(x)}\frac{2t}{x-1}dt=\displaystyle \lim_{x \to 1}\frac{1}{x-1}\int_{4}^{f(x)}2t\:dt=\displaystyle \lim_{x \to 1}\:\frac{1}{x-1}\left [t^2 \right ]_{4}^{f(x)}\end{array} \)
\(\begin{array}{l}=\displaystyle \lim_{x \to 1}\:\frac{\left\{ f(x)\right\}^{2}-4^{2}}{x-1}=\displaystyle \lim_{ x\to 1}[f(x)+4].\displaystyle \lim_{x \to 1}\frac{f(x)-4}{x-1}\end{array} \)
\(\begin{array}{l}=[f(1)+4].\displaystyle \lim_{x \to 1}\frac{f(x)-f(1)}{x-1}\:\:\:\left [ \because f(1)=4 \right ]\end{array} \)
\(\begin{array}{l}=8.\displaystyle \lim_{x \to 1}\frac{f(x)-f(1)}{x-1}\end{array} \)
Let x = 1 + h, where is a very small positive quantity such that when h → 0 ⇒ x → 1, then we have,
\(\begin{array}{l}=8.\displaystyle \lim_{h \to 0}\frac{f(1+h)-f(1)}{h}\end{array} \)
By the definition of derivative of a function at any point, we get
\(\begin{array}{l}\displaystyle \lim_{x \to 1}\int_{4}^{f(x)}\frac{2t}{x-1}dt=8f'(1).\end{array} \)
Video Lesson on Definite Integrals – Area Under the Curves
Practice Questions on Definite Integrals
1. Evaluate the following integrals:
\(\begin{array}{l}(i)\:\int_{0}^{\pi/4}\sqrt{1-sin\:2x}\:dx\end{array} \)
\(\begin{array}{l}(ii)\:\int _{0}^{1}x^{2}e^{x}\:dx\end{array} \)
\(\begin{array}{l}(iii)\:\int _{0}^{1}x.tan^{-1}x\:dx\end{array} \)
\(\begin{array}{l}(iv)\:\int_{1}^{5}\frac{log\:x}{(x+1)^{2}}dx\end{array} \)
2. Evaluate using the properties of definite integral:
\(\begin{array}{l}(i) \int_{0}^{2}x^{2}\sqrt{2-x}\:dx\end{array} \)
\(\begin{array}{l}(ii)\:\int_{0}^{2\pi}\frac{x.sin^{2n}x\:dx}{sin^{2n}x+cos^{2n}x},n\;\epsilon \;\mathbb{N}\end{array} \)
3. If
\(\begin{array}{l}\int_{a}^{b}x^{3}dx=0 \:and\:\int_{a}^{b}x^{2}dx=\frac{2}{3},\end{array} \)
find the value of a and b.
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