Free Definite Integrals and Areas 03 Practice Test - 12th Grade - Commerce

Question 1

Find the integral $\int_{0}^{\infty} e^{- x} d x$ .

A. 0

B. 1

C. 2

\infty

SOLUTION

Solution : B

We can see that the given integral is an improper integral as one of its limits is not finite.
In such cases where we have to deal with infinity as the limits of definite integral, we’ll change the limit which is not finite to a variable and then put the limits.
$\int_{0}^{\infty} e^{- x} d x . = l i m a \to \infty \int_{0}^{a} e^{- x} d x$
$= l i m a \to \infty (- e^{- x})_{0}^{a}$
$= lim a \to \infty (- e^{- a} - (- e^{0}))$
$= 0 - (- 1)$
$= 1$

Question 2

Area enclosed by curve $y^{3} - 9 y + x = 0$ and Y - axis is -

\frac{9}{2}

9

\frac{81}{2}

81

SOLUTION

Solution : C

The given equation of curve can be written as $x = f (y) = 9 y - y^{3}$ . Now to calculate the area we need to find the boundaries of this curve i.e ordinates or the point where this curve is meeting Y - axis.
$f (y) = y (9 - y^{2})$
$f (y) = y . (3 + y) (3 - y)$
So, the points where f(y) is meeting y - axis are y = -3, y = 0 & y = 3.
Important thing to note here is that the function is changing its signs.
I.e. from y = -3 to y = 0 f(y) is negative.
& from y = 0 to y = 3 f(y) is positive.
Since, the function in negative in the interval (-3, 0 ) we’ll take absolute value of it. Because we are interested in the area enclosed and not the algebraic sum of area.
Let the area enclosed be A.
$A = \int_{- 3}^{0} | f (y) | d y + \int_{0}^{3} f (y) d y A = \int_{- 3}^{0} | 9 y - y^{3} | d y + \int_{0}^{3} (9 y - y^{3}) d y A = - \int_{- 3}^{0} (9 y - y^{3}) d y + \int_{0}^{3} 9 y - y^{3} d y A = - [\frac{9 y^{2}}{2} - \frac{y^{4}}{4}]_{- 3}^{0} + [\frac{9 y^{2}}{2} - \frac{y^{4}}{4}]_{0}^{3} A = - [0 - 0 - \frac{81}{2} + \frac{81}{4}] + [\frac{81}{2} - \frac{81}{4}] A = 81 - \frac{81}{2} A = \frac{81}{2}$

Question 3

$\int_{0}^{10 π} | s i n x | d x i s$

A. 20

B. 8

C. 10

D. 18

SOLUTION

Solution : A

$∴ \int_{0}^{10 π} | s i n x | d x = 10 \int_{0}^{π} | s i n x | d x$
$∴ | s i n x | i s$ positive in I & II quadrant and has period $π$
$= 10 \int_{0}^{π} s i n x d x = 10 [- c o s x]_{0}^{x} = 20$

Question 4

$\int_{0}^{2} [x^{2}] d x$ is (where [.] is greastest integral function

2 - \sqrt{2}

2 + \sqrt{2}

\sqrt{2} - 1

- \sqrt{2} - \sqrt{3} + 5

SOLUTION

Solution : D

$\int_{0}^{2} [x^{2}] d x$
$= \int_{0}^{1} [x^{2}] d x + \int_{0}^{\sqrt{2}} [x^{2}] d x + \int_{\sqrt{2}}^{\sqrt{3}} [x^{2}] d x + \int_{\sqrt{3}}^{2} [x^{2}] d x$
$= \int_{0}^{1} 0 d x + \int_{0}^{\sqrt{2}} 1 d x + \int_{\sqrt{2}}^{\sqrt{3}} 2 d x + \int_{\sqrt{3}}^{2} 3 d x$
$= \sqrt{2} - 1 + 2 \sqrt{3} - 2 \sqrt{2} + 6 - 3 \sqrt{3}$
$= 5 - \sqrt{3} - \sqrt{2}$

Question 5

$\int_{0}^{\frac{π}{2}} l o g (t a n x + c o t x) d x =$

π l o g 2

- π l o g 2

- \frac{π}{2} l o g 2

\frac{π}{2} l o g 2

SOLUTION

Solution : A

$\int_{0}^{\frac{π}{2}} l o g (t a n x + c o t x) d x = \int_{0}^{\frac{π}{2}} l o g [\frac{2}{s i n 2 x}] d x$
$= \int_{0}^{\frac{π}{2}} (l o g 2 - l o g s i n 2 x) d x = l o g 2 (\frac{π}{2}) + (\frac{π}{2}) l o g 2 = π l o g 2$

Question 6

$\int_{0}^{1} \frac{x^{7}}{\sqrt{1 - x^{4}}} d x$ is equal to

A. 1

\frac{1}{3}

\frac{2}{3}

\frac{π}{3}

SOLUTION

Solution : B

$I = \int_{0}^{1} \frac{x^{7}}{\sqrt{1 - x^{4}}} d x = \int_{0}^{1} \frac{x^{6} x d x}{\sqrt{1 - x^{4}}}$
$P u t x^{2} = s i n θ \Rightarrow 2 x d x = c o s θ d θ$
$I = \frac{1}{2}_{0}^{\frac{π}{2}} \frac{s i n^{3} θ . c o s θ d θ}{c o s θ} = \frac{1}{2} \int_{0}^{\frac{1}{2}} s i n^{3} θ d θ$
$= \frac{1}{2} \frac{T 2 T (\frac{1}{2})}{2. T (\frac{1}{2})} = \frac{T (\frac{1}{2})}{4. \frac{3}{2} . \frac{1}{2} . T (\frac{1}{2})} = \frac{1}{3}$

Question 7

Let f be a positive function. Let
$I_{1} = \int_{1 - k}^{k} x f {x (1 - x)} d x, I_{2} = \int_{1 - k}^{k} f {x (1 - x)} d x$
$w h e n 2 k - 1 > 0. T h e n \frac{I_{1}}{I_{2}} i s$ [IIT 1997 Cancelled]

$\frac{1}{2}$

SOLUTION

Solution : C

$I_{1} = \int_{1 - k}^{k} x f {x (1 - x)} d x$
$= \int_{1 - k}^{k} (1 - k + k - x) f [(1 - k + k - x) {1 - (1 - k + k - x)}] d x$
$= \int_{1 - k}^{k} (1 - x) f {x (1 - x)} d x$
$= \int_{1 - k}^{k} f {x (1 - x)} d x - \int_{1 - k}^{k} x f {x (1 - x)} d x = I_{2} - I_{1}$
$∴ 2 I_{1} = I_{2} \Rightarrow \frac{I_{1}}{I_{2}} = \frac{1}{2}$

Question 8

If I is the greatest of the definite integrals
$I_{1} = \int_{0}^{1} e^{- x} c o s^{2} x d x, I_{2} = \int_{0}^{1} e^{- x^{2}} c o s^{2} x d x$
$I_{3} = \int_{0}^{1} e^{- x^{2}} d x, I_{4} = \int_{0}^{1} e^{\frac{- x^{2}}{2}} d x, t h e n$

$I = I_{1}$

$I = I_{2}$

$I = I_{3}$

$I = I_{4}$

SOLUTION

Solution : D

For 0 < x < 1, we have $\frac{1}{2} x^{2} < x^{2} < x$
$\Rightarrow - x^{2} > - x, s o t h a t e^{- x^{2}} < e^{- x},$
$H e n c e \int_{0}^{1} e^{- x^{2}} c o s^{2} x d x > \int_{0}^{1} e^{- x} c o s^{2} x d x$
$A l s o c o s^{2} x \leq 1$
$T h e r e f o r e \int_{0}^{1} e^{- x^{2}} c o s^{2} x d x \leq \int_{0}^{1} e^{- x^{2}} d x < \int_{0}^{1} e^{\frac{- x^{2}}{2}} d x = I_{4}$
Hence $I_{4}$ is the greatest integral

Question 9

Area bounded by parabola $y^{2} = x$ and straight line 2y = x is

\frac{4}{3}

1

$\frac{2}{3}$

$\frac{1}{3}$

SOLUTION

Solution : A

$y^{2} = x a n d 2 y = x \Rightarrow y^{2} = 2 y \Rightarrow y = 0, 2$
$∴ R e q u i r e d a r e a = \int_{0}^{2} (y^{2} - 2 y) d y = {(\frac{y^{3}}{3} - y^{2})}_{0}^{2} = \frac{4}{3} s q . u n i t$

Question 10

The correct evaluation of $\int_{0}^{π} | s i n^{4} x | d x$ is [MP PET 1993]

\frac{8 π}{3}

\frac{2 π}{3}

\frac{4 π}{3}

\frac{3 π}{8}

SOLUTION

Solution : D

$\int_{0}^{π} | s i n^{4} x | d x = 2 \int_{0}^{\frac{π}{2}} s i n^{4} x d x$
Applying gamma function,
$2 \int_{0}^{\frac{π}{2}} s i n^{4} x d x = 2 \frac{T (\frac{5}{2}) . T (\frac{1}{2})}{2. T (\frac{6}{2})} = \frac{3 π}{8}$

Free Definite Integrals and Areas 03 Practice Test - 12th Grade - Commerce

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

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