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Integrals

Multiple Choice Questions

31.

What is $\int_{- 2}^{2} x d x - \int_{- 2}^{2} [x] d x$

32.

If $\int_{- 2}^{5} f (x) d x = 4 and \int_{0}^{5} [1 + f (x)] d x = 7$

then what is $\int_{- 2}^{0} f (x) d x = ?$

33.

What is $\int_{0}^{4 π} |\cos (x)| dx$ = ?

34.

Let f(x) be a function such that $f' (\frac{1}{x}) + x^{3} f' (x) = 0 . What is \int_{- 1}^{1} f (x) d x = ?$

2f(1)
0
2f( - 1)
4f(1)

35.

What is $\int \frac{x^{4} - 1}{x^{2} \sqrt{x^{4} + x^{2} + 1}} dx = ?$

$\sqrt{\frac{x^{4} + x^{2} + 1}{4}} + c$
$\sqrt{x^{2} + 2 - \frac{1}{x^{2}}} + c$
$\sqrt{x^{2} + \frac{1}{x^{2}} + 1} + c$
$\sqrt{\frac{x^{4} - x^{2} + 1}{x}} + c$

36.

What is $\int e^{\sin (x)} \frac{{xcos}^{3} (x) - \sin (x)}{\cos^{2} (x)} dx = ?$

$(x + \sec (x)) e^{\sin (x)} + c$
$(x - \sec (x)) e^{\sin (x)} + c$
$(x + \tan (x)) e^{\sin (x)} + c$
$(x - \tan (x)) e^{\sin (x)} + c$

37.

If $\int_{0}^{\frac{π}{2}} \frac{dx}{3 \cos (x) 5} = {kcot}^{- 1} (2)$ , then what is the value of k ?

$\frac{1}{4}$
$\frac{1}{2}$
1
2

38.

What is $\int_{1}^{3} |1 - x^{4}| d x = ?$

$- \frac{232}{5}$
$- \frac{116}{5}$
$\frac{116}{5}$
$\frac{232}{5}$

39.
$Consider the integral I_{1} = \int_{0}^{π} \frac{x}{1 + \sin (x)} d x and I_{2} = \int_{0}^{π} \frac{(π - x) dx}{1 - \sin (π + x)}$
What is the value of I₁ + I₂ ?

2 $π$

$π$

$\frac{π}{2}$

0

2 $π$

$Consider the integral I_{1} = \int_{0}^{π} \frac{x}{1 + \sin (x)} d x Then, with the help of the property, we can write I_{1} = \int_{0}^{π} \frac{x}{1 + \sin (x)} d x . . . (i) As : I_{1} = \int_{0}^{π} \frac{(π + 0 - x) dx}{1 - \sin (π + 0 - x)} I_{1} = \int_{0}^{π} \frac{(π - x) dx}{1 + \sin (π - x)} I_{1} = \int_{0}^{π} \frac{(π - x)}{1 + \sin (x)} d x . . . (ii) Now, add equations (i) and (ii), we get 2 I_{1} = \int_{0}^{π} \frac{x}{1 + \sin (x)} d x 2 I_{1} = π \int_{0}^{π} \frac{1 - \sin (x)}{(1 + \sin (x)) (1 - \sin (x))} d x 2 I_{1} = π (\sec^{2} (x) - \sec (x) \tan (x)) dx 2 I_{1} = π {[\tan (x) - \sec (x)]}_{0}^{π} 2 I_{1} = [1 - (- 1)] 2 I_{1} = 2 π I_{1} = π Now I_{2} = \int_{0}^{π} \frac{(π - x) dx}{1 - \sin (π + x)} We will again the property of integration Then I_{2} = \int_{0}^{π} \frac{(π + 0 - (π - x)) dx}{1 - \sin (π + 0 - (π - x))} I_{2} = \int_{0}^{π} \frac{xdx}{1 - \sin (- x)} I_{2} = \int_{0}^{π} \frac{xdx}{1 + \sin (x)} = I_{1} = π Hence I_{1} + I_{2} = 2 π So option A is correct$