Free Definite Integrals and Areas 02 Practice Test - 12th Grade - Commerce

Question 1

$\int_{0}^{\frac{π}{2}} \frac{c o s x}{1 + s i n x} d x =$

log 2

log e

$\frac{1}{2} l o g 3$

SOLUTION

Solution : A

put 1+ sin x =t
$T h e n \int_{0}^{\frac{π}{2}} \frac{c o s x}{1 + s i n x} d x = {[l o g | 1 + s i n x |]}_{0}^{\frac{π}{2}} = l o g 2$

Question 2

If area bounded by the curves $y^{2} = 4 a x a n d y = m x i s \frac{a^{2}}{3}$ then the value of m is

A. 2

B. -2

$\frac{1}{2}$

D. None of these

SOLUTION

Solution : A

The two curves $y^{2}$ = 4ax and y = mx intersect at $(\frac{4 a}{m^{2}}, \frac{4 a}{m})$ and the area enclosed by the two curves is given by $\int_{0}^{\frac{4 a}{m^{2}}} (\sqrt{4 a x} - m x) d x$
$∴ \int_{0}^{\frac{4 a}{m^{2}}} (\sqrt{4 a x} - m x) d x = \frac{a^{2}}{3}$
$\Rightarrow \frac{8}{3} \frac{a^{2}}{m^{3}} = \frac{a^{2}}{3} \Rightarrow m^{3} = 8 \Rightarrow m = 2$

Question 3

$\int_{- \frac{π}{2}}^{\frac{π}{2}} s i n^{2} x d x =$

π

$\frac{π}{2}$

\frac{π}{2} - \frac{1}{2}

π - 1

SOLUTION

Solution : B

$\int_{- \frac{π}{2}}^{\frac{π}{2}} s i n^{2} x d x = 2 \int_{0}^{\frac{π}{2}} s i n^{2} x d x = 2 \frac{r (\frac{3}{2}) r (\frac{1}{2})}{2 r (\frac{2 + 2}{2})} = \frac{π}{2}$

Question 4

$\int_{0}^{\frac{π}{2}} s i n^{2} x c o s^{3} x d x =$ [RPET 1984, 2003]

\frac{2}{15}

\frac{4}{15}

None of these

SOLUTION

Solution : B

Using gamma function,
$\int_{0}^{\frac{π}{2}} s i n^{2} x c o s^{3} x d x = \frac{r (\frac{3}{2}) r 2}{2 r (\frac{7}{2})} = \frac{2}{15}$

Question 5

$\int_{0}^{\infty} \frac{l o g (1 + x^{2})}{1 + x^{2}} d x =$

$π l o g \frac{1}{2}$

π l o g 2

2 π l o g \frac{1}{2}

2 π l o g 2

SOLUTION

Solution : B

$L e t I = \int_{0}^{\infty} \frac{l o g (1 + x^{2})}{1 + x^{2}} d x$
$P u t x = t a n θ \Rightarrow d x = s e c^{2} θ d θ,$
$∴ I = \int_{0}^{\frac{n}{2}} l o g (s e c θ)^{2} d θ = 2 \int_{0}^{\frac{n}{2}} l o g s e c θ d θ$
$= - 2 \int_{0}^{\frac{n}{2}} l o g c o s θ d θ$ $= - 2. \frac{π}{2} l o g \frac{1}{2} = - π l o g \frac{1}{2} = π l o g 2$

Question 6

${lim}_{π \to \infty} \frac{1^{99} + 2^{99} + 3^{99} + \dots \dots n^{99}}{n^{100}} =$ [EAMCET 1994]

$\frac{9}{100}$

$\frac{1}{100}$

$\frac{1}{99}$

$\frac{1}{101}$

SOLUTION

Solution : B

${lim}_{π \to \infty} \frac{1^{99} + 2^{99} + 3^{99} + \dots \dots n^{99}}{n^{100}} = {lim}_{π \to \infty} \sum_{r = 1}^{n} (\frac{r^{99}}{n^{100}})$
$= {lim}_{π \to \infty} \frac{1}{n} \sum_{r = 1}^{n} {(\frac{r}{n})}^{99} = \int_{0}^{1} x^{99} d x = {[\frac{x^{100}}{100}]}_{0}^{1} = \frac{1}{100}$

Question 7

$l i m_{π \to \infty} \sum_{r = 1}^{n} \frac{1}{n} e^{\frac{r}{n}} i s$ [AIEEE 2004]

A. e + 1

B. e - 1

C. 1 - e

D. e

SOLUTION

Solution : B

$l i m_{π \to \infty} \sum_{r = 1}^{n} \frac{1}{n} e^{\frac{r}{n}} = \int_{0}^{1} e^{x} d x = [e^{x}]_{0}^{1} = e - 1$

Question 8

The value of the definite integral $\int_{0}^{1} \frac{x d x}{x^{3} + 16}$ lies in the interval [a, b]. The smallest such interval is.

[0, \frac{1}{17}]

$[0, 1]$

[0, \frac{1}{27}]

$N o n e o f t h e s e$

SOLUTION

Solution : A

$f (x) = \frac{x}{x^{3} + 16} \Rightarrow f^{'} (x) = \frac{16 - 2 x^{3}}{(x^{3} + 16)^{2}} ∴ f^{'} (x) = 0 \Rightarrow x = 2 Also f^{''} (2) < 0 \Rightarrow$ The function $f (x) = \frac{x}{x^{3} + 16}$ is an increasing function in [0,1], so Min f(x) = f(0) = 0 and Max f(x) = f(1) = $\frac{1}{17}$
Therefore by the property $m (b - a) \leq \int_{a}^{b} f (x) d x \leq M (b - a)$
(where m and M are the smallest and greatest values of function)
$\Rightarrow 0 \leq \int_{0}^{1} \frac{x}{x^{3} + 16} d x \leq \frac{1}{17}$

Question 9

If $\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t$ , then the value of f(1) is [IIT 1998; AMU 2005]

\frac{1}{2}

C. 1

$- \frac{1}{2}$

SOLUTION

Solution : A

$\int_{0}^{x} d t = x + \int_{x}^{1} t f (t) d t \Rightarrow \int_{x}^{1} t f (t) d t = x - \int_{1}^{x} t f (t) d t$
Differentiating w.r.t x, we get $f (x) = 1 + {0 - x f (x)}$
$\Rightarrow f (x) = 1 - x f (x) \Rightarrow (1 + x) f (x) = 1 \Rightarrow f (x) = \frac{1}{1 + x}$
$∴ f (1) = \frac{1}{1 + 1} = \frac{1}{2}$

Question 10

Calculate $\int_{- 2}^{3} f (x) d x$ where
$f (x) = {\begin{matrix} 6 & i f x > 1 3 x^{2} & i f x \leq 1 \end{matrix}$

A. 19

B. 21

C. 17

D. 15

SOLUTION

Solution : B

Here, we can see that the integrand f(x) in not continuous in the interval (-2, 3) as it has a discontinuity at x = 1. So we can’t just integrate f(x) and put limits.We’ll have to break this integral into integrals which have limits such that integrands is continuous in those limits.
$\int_{- 2}^{3} f (x) d x = \int_{- 2}^{1} 3 x^{2} d x + \int_{1}^{3} 6 d x$
Now we can integrate these integrands and put limits.
$\int_{- 2}^{1} 3 x^{2} d x + \int_{1}^{3} 6 d x = (x^{3}) |_{- 2}^{1} + (6 x) |_{1}^{3} = 1 - (- 8) + 18 - 6$
= 21

Free Definite Integrals and Areas 02 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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