Calculus

Integrals

Easy

core

Match the LIST-I with LIST-II

LIST-I
Definite Integrals LIST-II
Values
A. $\displaystyle\int_{-1}^{1} \vert x \vert \, dx$ I. $2$
B. $\displaystyle\int_{-1}^{1} \sin^3 x \cos^4 x \, dx$ II. $1$
C. $\displaystyle\int_{-1}^{1} 3\vert x \vert^2 \, dx$ III. $0$
D. $\displaystyle\int_{0}^{4} \vert x-1 \vert \, dx$ IV. $5$

Choose the correct answer from the options given below:

	LIST-I Definite Integrals		LIST-II Values
A.	$\displaystyle\int_{-1}^{1} \vert x \vert \, dx$	I.	$2$
B.	$\displaystyle\int_{-1}^{1} \sin^3 x \cos^4 x \, dx$	II.	$1$
C.	$\displaystyle\int_{-1}^{1} 3\vert x \vert^2 \, dx$	III.	$0$
D.	$\displaystyle\int_{0}^{4} \vert x-1 \vert \, dx$	IV.	$5$

A-II, B-III, C-I, D-IV

A-IV, B-I, C-II, D-III

A-III, B-II, C-IV, D-I

A-II, B-I, C-IV, D-III

✅ Correct Option: 1

A: $\int_{-1}^1|x|dx = 2\int_0^1 x\,dx = 1$ (II).

B: the integrand is odd, so the value is $0$ (III).

C: $3\int_{-1}^1 x^2dx = 3(2/3) = 2$ (I).

D: $\int_0^1(1-x)dx + \int_1^4(x-1)dx = 0.5 + 4.5 = 5$ (IV).

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If $f'(x) = 4x^5 - 6x$ and $f(0) = 3$ , then $f(3)$ is equal to

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Value of $\int_{1}^{4} \frac{\log_e x}{x^2}\, dx$ is

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Match List I with List II

LIST I	LIST II
A. $\int \frac{\sin x}{1 + \cos x} \, dx$	I. $e^{\tan^{-1} x} + C$
B. $\int \frac{1}{1 - \tan x} \, dx$	II. $\log(\log x + 1) + C$
C. $\int \frac{e^{\tan^{-1} x}}{1 + x^2} \, dx$	III. $-\log\lvert 1+\cos x \rvert + C$
D. $\int \frac{1}{x + x \log x} \, dx$	IV. $\frac{x}{2} - \frac{1}{2}\log\lvert \cos x - \sin x \rvert + C$

Choose the correct answer from the options given below:

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$\int \frac{\cos 2x - \cos 2α}{\cos x - \cos α} dx$ is equal to

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The value of the integral $\int_{-6}^{0} |x+3|\,dx$ is

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The integral $I = \int \frac{(1+\log x)^3}{x}\, dx$ equals:

Solution