3 Marks Question

Question 13 Marks

Prove that $\int_a^b f(a+b-x) d x=\int_a^b f(x) d x$ and using it find $\int_4^9 \frac{f(x)}{f(x)+f(13-x)} d x$.

Answer

Let $\quad I=\int_a^b f(a+b-x) d x \quad \quad \ldots \ldots(1)$
$t=a+b-x$, then $d t=-d x \Rightarrow d x=-d t$
When $x=a$, then $t=b, x=b$ then $t=a$
Substituting $I=\int_b^a f(t)(-d t)$
$\begin{array}{ll}
\quad=-\int_b^a f(t) d t & \\
\quad=\int_b^a f(t) d t & \left(\because-\int_a^b f(x) d x=\int_b^a f(x) d x\right) \\
\quad=\int_b^a f(x) d x & \left(\because \int_a^b f(x) d x=\int_b^a f(y) d y\right)
\end{array}$
Here $\int_b^a f(a+b-x) d x=\int_b^a f(x) d x$
Hence proved.
Again let $I _1=\int_4^9 \frac{f(x)}{f(x)+f(13-x)} d x \quad \quad \ldots \ldots(2)$
Using above properties :
$
\begin{aligned}
I_1 & =\int_4^9 \frac{f(4+9-x)}{f(4+9-x)+f\{13-(4+9-x)\}} d x \\
& =\int_4^9 \frac{f(13-x)}{f(13-x)+f(13-13+x)} d x \\
I_1 & =\int_4^9 \frac{f(13-x)}{f(13-x)+f(x)} d x \quad \quad \ldots \ldots(3)
\end{aligned}
$
Adding eqns (2) and (3),
$
\begin{aligned}
2 I_1 & =\int_4^9 \frac{f(x)+f(13-x)}{f(13-x)+f(x)} d x \\
& =\int_4^9 1 \cdot d x \\
& =(x)_4^9=(9-4)=5
\end{aligned}
$
So, $\quad I _1=\frac{5}{2}$
$\Rightarrow \int_4^9 \frac{f(x)}{f(x)+f(13-x)} d x=\frac{5}{2}$

View full question & answer→

Question 23 Marks

Find the value of $\int \sqrt{5-4 x-x^2} d x$.

Answer

$\int \sqrt{5-4 x-x^2} d x $
$=\int \sqrt{-\left(x^2+4 x-5\right)} d x$
$=\int \sqrt{-\left(x^2+4 x+4-9\right)} d x $
$=\int \sqrt{-\left((x+2)^2-9\right)} d x$
$ =\int \sqrt{9-(x+2)^2} d x$
Putting
$(x+2)=t$
$ \Rightarrow d x=d t$
$=\int \sqrt{9-t^2} d t$
$=\int \sqrt{(3)^2-(t)^2} d t$
$=\frac{t}{2} \sqrt{9-t^2}+\frac{9}{2} \sin ^{-1}\left(\frac{t}{3}\right)+C$
${\left(\because \int \sqrt{a^2-x^2} d x\right).}$
$=\frac{1}{2} x \sqrt{a^2-x^2}+\frac{1}{2} a^2$
$\left.\sin ^{-1}\left(\frac{x}{a}\right)+C\right)$
$=\frac{(x+2)}{2} \sqrt{9-(x+2)^2}+$
$\frac{9}{2} \sin ^{-1}\left(\frac{x+2}{3}\right)+C$
$=\frac{(x+2)}{2} \sqrt{5-4 x-x^2}+$
$\frac{9}{2} \sin ^{-1}\left(\frac{x+2}{3}\right)+C \text { }$

View full question & answer→

Question 33 Marks

Integrate the function $\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}$ with respect to $x$.

Answer

Let
$I=\int \frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}} d x$
$I=\int \frac{\sin ^{-1} x}{\left(1-x^2\right) \sqrt{1-x^2}} d x$

$\text { Putting } \sin ^{-1} x=t $
$\Rightarrow \frac{1}{\sqrt{1-x^2}} d x=d t$
$I=\int \frac{t d t}{\left(1-\sin ^2 t\right)}=\int \frac{t d t}{\cos ^2 t}=\int t \sec ^2 t d t$
$I=\int_{I}^{t II} t \sec ^2 t d t$
$($using $\text{ILATE})$
$I=t \cdot \tan t-\int 1 \cdot \tan t d t$
$=t \tan t-\log \sec t+C$
$\therefore I=\sin ^{-1} x \tan \left(\sin ^{-1} x\right)-\log \sec \left(\sin ^{-1} x\right)+C$
$I=\sin ^{-1} x \tan \left(\tan ^{-1} \frac{x}{\sqrt{1-x^2}}\right)-$
$\log \sec \left(\sec ^{-1} \frac{1}{\sqrt{1-x^2}}\right)+C$
$=\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}-\log \frac{1}{\sqrt{1-x^2}}+C$
$=\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}-\log \left(1-x^2\right)^{-\frac{1}{2}}+C$
$=\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\frac{1}{2} \log \left(1-x^2\right)+C$

View full question & answer→

Question 43 Marks

Integrate the function $\frac{x}{1+\sin x}$ with respect to $x$.

Answer

Let $\quad I =\frac{x}{1+\sin x} d x$
On multiplying by $(1-\sin x)$ in numerator and denominator,
$
\begin{aligned}
I & =\int \frac{x(1-\sin x)}{(1+\sin x)(1-\sin x)} d x \\
& =\int \frac{x(1-\sin x)}{1-\sin ^2 x} d x=\int \frac{x(1-\sin x)}{\cos ^2 x} d x \\
& =\int_{\text {I }}^x \sec _{\text {II }}{ }^2 x d x-\int_{I} x \sec x \tan x d x \quad \text { I (using ILATE) }\\ & =x \cdot \tan x-\int 1 \cdot \tan x d x-\quad\left(x \cdot \sec x-\int 1 \cdot \sec x d x\right) \\
& =x \tan x-\int \tan x d x-x \sec x+\int \sec x d x \\
& =x \tan x-\log \sec x-x \sec x \\
& +\log (\sec x+\tan x)+C \\
& =x(\tan x-\sec x)+\log \left(\frac{\sec x+\tan x}{\sec x}\right)+C
\end{aligned}
$

View full question & answer→

MATHS 3 Marks Question

Take a timed test