Definite Integration(11th And 12th > Mathematics ) Questions and answers for exam Preparation

11th And 12th > Mathematics

DEFINITE INTEGRATION MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1.

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

log 2
log e
$\frac{1}{2} l o g 3$
0

Discuss Question

Answer: Option A. -> log 2
:
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.

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

$π$
$\frac{π}{2}$
$\frac{π}{2} - \frac{1}{2}$
$π - 1$

Discuss Question

Answer: Option B. ->

\frac{π}{2}

:
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 3.

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

2
-2
$\frac{1}{2}$
None of these

Discuss Question

Answer: Option A. -> 2
:
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 4.

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

0
$\frac{2}{15}$
$\frac{4}{15}$
None of these

Discuss Question

Answer: Option B. ->

\frac{2}{15}

:
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.

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

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

Discuss Question

Answer: Option B. ->

π l o g 2

:
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]

$\frac{9}{100}$
$\frac{1}{100}$
$\frac{1}{99}$
$\frac{1}{101}$

Discuss Question

Answer: Option B. ->

\frac{1}{100}

:
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.

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.

$[0, \frac{1}{17}]$
$[0, 1]$
$[0, \frac{1}{27}]$
$N o n e o f t h e s e$

Discuss Question

Answer: Option A. ->

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

:
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 8.

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

e + 1
e - 1
1 - e
e

Discuss Question

Answer: Option B. -> e - 1
:
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 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]

$\frac{1}{2}$
0
1
$- \frac{1}{2}$

Discuss Question

Answer: Option A. ->

\frac{1}{2}

:
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.

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

$0$
$1$
$2$
$\infty$

Discuss Question

Answer: Option B. ->

1

:
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.