Integral Calculus - Definite Integrals (Lecture-01)
If $f(\mathrm{x})$ is continuous on $[\mathrm{a}, \mathrm{b}]$ and $\mathrm{F}$ is an antiderivative (primitive) of $f$.
i.e. $f(\mathrm{x})=\dfrac{\mathrm{d}}{\mathrm{dx}} \mathrm{F}$, then
$\left.\int _{\mathrm{a}}^{\mathrm{b}} f(\mathrm{x}) \mathrm{dx}=\mathrm{F}(\mathrm{x})\right] _{\mathrm{a}}^{\mathrm{b}}=\mathrm{F}(\mathrm{b})-\mathrm{F}(\mathrm{a})$
Note: $\int _{a}^{a} f(x) d x=0$
Geometrical interpretation of definite integral
If $f(\mathrm{x})>0 \forall \mathrm{x} \in[\mathrm{a}, \mathrm{b}]$, then $\int _{\mathrm{a}}^{\mathrm{b}} f(\mathrm{x}) \mathrm{dx}$ is numerically equal to the area bonded by the curve $\mathrm{y}=f(\mathrm{x})$, the $\mathrm{x}$ - axis and the straight lines $\mathrm{x}$ = $\mathrm{a}$ & $\mathrm{x}$ = $\mathrm{b}$.
Properties of definite integrals
1. $\int _{\mathrm{a}}^{\mathrm{b}} f(\mathrm{x}) \mathrm{dx}=\int _{\mathrm{a}}^{\mathrm{b}} f(\mathrm{t}) \mathrm{dt}$
2. $\int _{\mathrm{a}}^{\mathrm{b}} f(\mathrm{x}) \mathrm{d} \mathrm{x}=-\int _{\mathrm{b}}^{\mathrm{a}} f(\mathrm{x}) \mathrm{d} \mathrm{x}$
3. $\int _{\mathrm{a}}^{\mathrm{b}} f(\mathrm{x}) \mathrm{dx}=\int _{\mathrm{a}}^{\mathrm{c}} f(\mathrm{x}) \mathrm{dx}+\int _{\mathrm{c}}^{\mathrm{b}} f(\mathrm{x}) \mathrm{dx}$ where $\mathrm{c} \in \mathrm{R}$
4. $\int _{\mathrm{a}}^{\mathrm{b}} f(\mathrm{x}) \mathrm{dx}=\int _{\mathrm{a}}^{\mathrm{b}} f(\mathrm{a}-\mathrm{x}) \mathrm{dx}$
5. $\int _{\mathrm{a}}^{\mathrm{b}} f(\mathrm{x}) \mathrm{dx}=\int _{\mathrm{a}}^{\mathrm{b}} f(\mathrm{a}+\mathrm{b}-\mathrm{x}) \mathrm{dx}$
6. $\int _{-a}^a f(\mathrm{x}) \mathrm{d} x=\int _0^{\mathrm{a}}(f(\mathrm{x})+f(-\mathrm{x})) \mathrm{d} \mathrm{x}$ = $\begin{cases}{l} 2 \int _0^a f(\mathrm{x}) \mathrm{d} x \text { if } f(-\mathrm{x})=f(\mathrm{x}) \\ \text { ie. if } f(\mathrm{x}) \text { is even } \\ 0, \text { if } f(-\mathrm{x})=-f(\mathrm{x}) \\ (\text { ie. if } f(\mathrm{x}) \text { is odd }) \end{cases}$.
7. $\int _{\mathrm{a}}^{2 \mathrm{a}} f(\mathrm{x}) \mathrm{dx}=\int _{0}^{\mathrm{a}}(f(\mathrm{x})+f(2 \mathrm{a}-\mathrm{x})) \mathrm{dx}$ = $\begin{cases}2 \int _{0}^{\mathrm{a}} f(\mathrm{x}) \mathrm{dx} \text { if } f(2 \mathrm{a}-\mathrm{x})=f(\mathrm{x}) \\ 0 \quad \text { ,if } f(2 \mathrm{a}-\mathrm{x})=-f(\mathrm{x})\end{cases}$.
8. $\int _{b}^{a} f(x) d x=(b-a) \int _{0}^{1} f((b-a) x+a) d x$
9. If $\mathrm{m}$ & $\mathrm{M}$ be the least and greatest values of $f(\mathrm{x})$ respectively on $[\mathrm{a}, \mathrm{b}]$, then $\mathrm{m}(\mathrm{b}-\mathrm{a}) \leq \int _{\mathrm{a}}^{\mathrm{b}} f(\mathrm{x}) \mathrm{dx} \leq \mathrm{M}$.
10. $\int _{0}^{1} \mathrm{x}^{m}(1-\mathrm{x})^{n} \mathrm{dx}=\dfrac{m ! n !}{(m+n+1) !}$
11. $\int _{\mathrm{a}}^{\mathrm{b}}(f \circ g)(\mathrm{x}) \mathrm{g}^{1}(\mathrm{x}) \mathrm{dx}=\int _{\mathrm{a}}^{\mathrm{b}} f(\mathrm{~g}(\mathrm{x})) \mathrm{g}^{1}(\mathrm{x}) \mathrm{dx}=\int _{\mathrm{g}(\mathrm{a})}^{\mathrm{g}(\mathrm{b})} f(\mathrm{t}) \mathrm{dt}$
12. If $f(\mathrm{x})$ is a periodic function with period $\mathrm{T}$ then
i. $\int _{0}^{\mathrm{nT}} f(\mathrm{x}) \mathrm{dx}=\mathrm{n} \int _{0}^{\mathrm{T}} f(\mathrm{x}) \mathrm{dx} \mathrm{n} \in \mathrm{Z}$
ii. $\int _{a}^{a+n T} f(x) d x=n \int _{a}^{a+T} f(x) d x=n \int _{0}^{T} f(x) d x$
which is independent of a
iii. $\int _{a+n \mathrm{~T}}^{b+n \mathrm{~T}} f(\mathrm{x}) \mathrm{dx}=\int _{\mathrm{a}}^{\mathrm{b}} f(\mathrm{x}) \mathrm{dx} \quad, \mathrm{n} \in \mathrm{Z}$
iv. $\int _{\mathrm{mT}}^{\mathrm{nT}} f(\mathrm{x}) \mathrm{dx}=(\mathrm{n}-\mathrm{m}) \int _{0}^{\mathrm{T}} f(\mathrm{x}) \mathrm{dx}, \mathrm{m}, \mathrm{n} \in \mathrm{Z}$.
v. If $f(\mathrm{t})$ is an odd (even) function periodic with period $\mathrm{T}$, then $\mathrm{F}(\mathrm{x})=\int _{\mathrm{a}}^{\mathrm{x}} f(\mathrm{t}) \mathrm{dt}$ is an even (odd) function with period $\mathrm{T}$.
13. If $f(\mathrm{x}) \leq \mathrm{g}(\mathrm{x})$ on $[\mathrm{a}, \mathrm{b}]$, then $\int _{\mathrm{a}}^{\mathrm{b}} f(\mathrm{x}) \mathrm{dx} \leq \int _{\mathrm{a}}^{\mathrm{b}} \mathrm{g}(\mathrm{x}) \mathrm{dx}$.
Also $\int _{\mathrm{a}}^{\mathrm{b}} f(\mathrm{x}) \mathrm{d} \mathrm{x} \leq \int _{\mathrm{a}}^{\mathrm{b}}|f(\mathrm{x})| \mathrm{d} \mathrm{x}$
14. Leibnitz Rule
i. $\dfrac{\mathrm{d}}{\mathrm{dx}} \int _{\varphi(\mathrm{x})}^{\Psi(\mathrm{x})} f(\mathrm{t}) \mathrm{dt}=f(\Psi(\mathrm{x})) \dfrac{\mathrm{d}}{\mathrm{dx}}(\Psi(\mathrm{x}))-f(\varphi(\mathrm{x})) \dfrac{\mathrm{d}}{\mathrm{dx}}(\varphi(\mathrm{x}))$
ii. $\dfrac{\mathrm{d}}{\mathrm{dx}} \int _{\varphi(\mathrm{x})}^{\Psi(\mathrm{x})} f(\mathrm{x}, \mathrm{t}) \mathrm{dt}=\int _{\varphi(\mathrm{x})}^{\Psi(\mathrm{x})} \dfrac{\partial f(\mathrm{x}, \mathrm{t})}{\partial \mathrm{x}} \mathrm{dt}+\dfrac{\mathrm{d} \Psi(\mathrm{x})}{\mathrm{dx}} f(\mathrm{x}, \Psi(\mathrm{x}))-\dfrac{\mathrm{d} _{\varphi}(\mathrm{x})}{\mathrm{dx}} f(\mathrm{x}, \varphi(\mathrm{x}))$
Reduction Formulae
$\quad\mathrm{I} _{\mathrm{n}}=\int _{0}^{\pi / 2} \sin ^{\mathrm{n}} \mathrm{xdx}=\int _{0}^{\pi / 2} \cos ^{\mathrm{n}} \mathrm{xdx}=\dfrac{\mathrm{n}-1}{\mathrm{n}} \mathrm{I} _{\mathrm{n}-2}, \mathrm{n} \in \mathrm{Z}^{+}$
1. = $\begin{cases}\left(\dfrac{n-1}{n}\right)\left(\dfrac{n-3}{n-2}\right) \ldots \ldots . . \dfrac{1}{2} \cdot \dfrac{\pi}{2}, \text { If } n \text { is even } \\ & \\ \left(\dfrac{n-1}{n}\right)\left(\dfrac{n-3}{n-2}\right) \ldots \ldots \ldots \dfrac{2}{3}, \text { If } n \text { is odd. }\end{cases}$.
2. $\mathrm{I} _{\mathrm{n}}=\int _{0}^{\pi / 4} \tan ^{\mathrm{n}} \mathrm{x} d \mathrm{x} \quad \mathrm{n}>1$,
$\mathrm{I} _{\mathrm{n}}+\mathrm{I} _{\mathrm{n}-2}=\dfrac{1}{\mathrm{n}-1}$
$I _{n}=\dfrac{1}{n-1}-\dfrac{1}{n-3}+\dfrac{1}{n-5}-\dfrac{1}{n-7} \ldots \ldots \ldots \ldots \ldots I _{1} \text { or } I _{0}$
according as $\mathrm{n}$ is odd or even.
$\left(\mathrm{I} _{0}=\dfrac{\pi}{4}, \mathrm{I},=\dfrac{1}{2} \log _{\mathrm{e}} 2=\log _{\mathrm{e}} \sqrt{2}\right)$
iii. $\quad \int _{0}^{\pi / 4}\left(\tan ^{n} x+\tan ^{n-2} x\right) d x=\int _{\pi / 4}^{\pi / 2}\left(\cot ^{n} x+\cot ^{n-2} x\right) d x=\dfrac{1}{n-1}$
Walli’s Formulae.
$\begin{aligned} & I _{m, n}=\int _{0}^{\pi / 2}\left(\sin ^{m} x \cos ^{n} x\right) d x \\ \\ & =\dfrac{m-1}{m+1} I _{m-2, n} \\ \\ & =\dfrac{n-1}{m+1} I _{m, n-2} \text { where } m, n \in Z^{x} \\ \\ & I _{m, n}=\int _{0}^{\pi / 2}\left(\sin ^{m} x \cos ^{n} x\right) d x \\ \\ & =\dfrac{((m-1)(m-3) \ldots \ldots .1 \text { or } 2)((n-1)(n-3) \ldots \ldots . .1 \text { or } 2)}{(m+n)(m+n-2)(m+n-4) \ldots \ldots . .(1 \text { or } 2)} k \end{aligned}$
where $\mathrm{k}=\dfrac{\pi}{2}$ if both $\mathrm{m}$ & $\mathrm{n}$ are even
$\mathrm{k}=1 \quad \text { otherewise. }$
Improper integral
If $f(\mathrm{x})$ is continuous on $[\mathrm{a}, \infty)$ then $\int _{\mathrm{a}}^{\infty} f(\mathrm{x}) \mathrm{dx}=\lim\limits _{\mathrm{b} \rightarrow \infty} \int _{\mathrm{a}}^{\mathrm{b}} f(\mathrm{x}) \mathrm{dx}$ is called an improper integral.
If there exists a finite limit on right side of the above equation, we say the improper integral is convergent, otherwise it is divergent.
Similarity $\int _{-\infty}^{b} f(x) d x=\lim\limits _{\mathrm{a} \rightarrow-\infty} \int _{\mathrm{a}}^{\mathrm{b}} f(\mathrm{x}) \mathrm{dx}$
$\hspace{1.3cm}\int _{-\infty}^{\infty} f(\mathrm{x}) \mathrm{dx}=\int _{-\infty}^{\mathrm{a}} f(\mathrm{x}) \mathrm{dx}+\int _{\mathrm{a}}^{\infty} f(\mathrm{x}) \mathrm{dx}$
Geometrically, for $f(\mathrm{x})>0, \int _{\mathrm{a}}^{\infty} f(\mathrm{x}) \mathrm{dx}$ gives area of the figure bounded by $\mathrm{y}=f(\mathrm{x}), \mathrm{x}-$ axis & $\mathrm{x}=\mathrm{a}$.
Gamma Function
If $n$ is a positive rational number, then the improper integral $\int _{0}^{\infty} e^{-\mathrm{x}} \mathrm{x}^{n-1} \mathrm{dx}, \mathrm{x} \in Q^{+}$ is defined as
Gamma function and is denoted by $\cdot \Gamma \mathrm{n}$
$\therefore \Gamma \mathrm{n}=\int _{0}^{\infty} \mathrm{e}^{-\mathrm{x}} \mathrm{x}^{\mathrm{n}-1} \mathrm{dx}, \mathrm{x} \in \mathrm{Q}^{+}$
Properties
i. $\Gamma \mathrm{n}+1=\mathrm{n} \Gamma \mathrm{n}=\mathrm{n} ! ; \mathrm{n} \in \mathrm{N}$
ii. $\Gamma 1=1$ and $\Gamma 0=\infty, \quad \Gamma _{\dfrac{1}{2}}=\sqrt{\pi}$
iii. $\int _{0}^{\pi / 2} \sin ^{\mathrm{m}} x \cos ^{\mathrm{n}} \mathrm{xdx}=\dfrac{\dfrac{\Gamma _{\mathrm{m}+1}}{2} \dfrac{\Gamma _{\mathrm{n}+1}^{2}}{2}}{2 \dfrac{\sqrt[\mathrm{m}+\mathrm{n}+2]{2}}{2}}$
iv. $\Gamma{\mathrm{n}} \Gamma \mathrm{n}-1$ = $\dfrac{\pi}{\sin \mathrm{n} \pi}, 0<\mathrm{n}<1$.
v. $\Gamma{\mathrm{n}} \Gamma _{\mathrm{m}+\dfrac{1}{2}}=\dfrac{\sqrt{\pi}}{2^{2 \mathrm{~m}-1}} \Gamma 2 \mathrm{~m}$
vi. $\Gamma _{\dfrac{1}{n}} \Gamma _{\dfrac{2}{n}} \Gamma _{\dfrac{3}{n}} \cdots . . \Gamma _{\dfrac{n-1}{n}}=\dfrac{(2 \pi)^{\dfrac{n-1}{2}}}{n^{\dfrac{1}{2}}}$
Beta Function
The Beta function is denoted by $B(m, n)$ where $B(m, n)=\int _{0}^{1} x^{m-1}(1-x)^{n-1} d x, m, n>0$
Properties
$\begin{array}{lll} \text{i. }& B(m, n) & =\int _{0}^{1} x^{m-1}(1-x)^{n-1} d x, m, n>0 \\ \\ & & =\int _{0}^{1} \dfrac{x^{m-1}}{(1+x)^{m+n}} d x \\ \\ & & =\dfrac{\Gamma \mathrm{m} \Gamma \mathrm{n}}{\Gamma(\mathrm{m}+\mathrm{n})} \end{array}$
Integration as limit of a sum
$\int _{0}^{1} f(\mathrm{x}) \mathrm{dx}=\lim \limits _{\mathrm{n} \rightarrow \infty} \dfrac{1}{\mathrm{n}} \sum\limits _{\mathrm{r}=1}^{\mathrm{n}} f\left(\dfrac{\mathrm{r}-1}{\mathrm{n}}\right)=\lim\limits _{\mathrm{n} \rightarrow \infty} \dfrac{1}{\mathrm{n}} \sum\limits _{\mathrm{r}=0}^{\mathrm{n}-1} f\left(\dfrac{\mathrm{r}}{\mathrm{n}}\right)$
Here replace $\dfrac{\mathrm{r}}{\mathrm{n}}$ by $x$
replace $\dfrac{1}{\mathrm{n}}$ by $\mathrm{dx}$
and $\lim\limits _{n \rightarrow \infty} \sum$ by $\int$
Solved Examples
1. If $\int _{0}^{1} \dfrac{e^{t} d t}{t+1}=a$, then $\int _{b-1}^{b} \dfrac{e^{-t} d t}{t-b-1}$ is equal to
(a) $\mathrm{ae}^{-\mathrm{b}}$
(b) $-\mathrm{ae}^{-\mathrm{b}}$
(c) $-b e^{-a}$
(d) $a e^{b}$
Show Answer
Solution: $\int _{b-1}^{b} \dfrac{e^{-t} d t}{t-b-1}$ $\hspace {2 cm}$ $\begin{array}{cl} \text { Put } & t-b=-y \\ & d t=-d y \end{array}$
$\Rightarrow-\int _{1}^{0} \dfrac{\mathrm{e}^{\mathrm{y}-\mathrm{b}}}{-\mathrm{y}-1} \mathrm{dy}=-\int _{0}^{1} \dfrac{\mathrm{e}^{-b} \mathrm{e}^{\mathrm{y}} \mathrm{dy}}{(\mathrm{y}+1)}=-\mathrm{ae}^{-\mathrm{b}}$
Answer: b
2. If $\int _{0}^{\infty} \mathrm{e}^{-\mathrm{x}^{2}} \mathrm{dx}=\dfrac{\sqrt{\pi}}{2}$, then $\int _{0}^{\infty} \mathrm{e}^{-\mathrm{ax}} \mathrm{dx}(\mathrm{a}>0)$ is equal to
(a) $\dfrac{\sqrt{\pi}}{2}$
(b) $\dfrac{\sqrt{\pi}}{2 \mathrm{a}}$
(c) $\dfrac{2 \sqrt{\pi}}{\mathrm{a}}$
(d) $\dfrac{\sqrt{\pi}}{2 \sqrt{\mathrm{a}}}$
Show Answer
Solution: If $\int _{0}^{\infty} \mathrm{e}^{-a x^{2}} \mathrm{dx} ; \quad \quad$ Put $\mathrm{ax}^{2}=\mathrm{t}^{2}$
$\hspace {4 cm}\Rightarrow 2 \mathrm{axdx}=2 \mathrm{tdt}$
$\therefore \int _{0}^{\infty} \dfrac{\mathrm{e}^{-t^{2}} \mathrm{tdt}}{\mathrm{a} \dfrac{\mathrm{t}}{\sqrt{\mathrm{a}}}}=\dfrac{1}{\mathrm{a}} \int _{0}^{\infty} \dfrac{\mathrm{e}^{-\mathrm{t}^{2}} \mathrm{tdt}}{\sqrt{\mathrm{a}} \mathrm{t}}$
$=\dfrac{1}{\sqrt{\mathrm{a}}} \int _{0}^{\infty} \mathrm{e}^{-\mathrm{t}^{2}} \mathrm{dt}=\dfrac{1}{\sqrt{\mathrm{a}}} \int _{0}^{\infty} \mathrm{e}^{-\mathrm{x}^{2}} \mathrm{dx}=\dfrac{\sqrt{\pi}}{2 \sqrt{\mathrm{a}}}$
Answer: d
3. Given $\int _{1}^{2} \mathrm{e}^{\mathrm{x}^{2}} \cdot \mathrm{dx}=\mathrm{a}$, then the value of $\int _{\mathrm{e}}^{\mathrm{e}^{4}} \sqrt{\log _{\mathrm{e}} \mathrm{x}} \mathrm{dx}$ is equal to
(a) $\mathrm{e}^{4}-\mathrm{e}$
(b) $\mathrm{e}^{4}-\mathrm{a}$
(c) $2 \mathrm{e}^{4}-\mathrm{a}$
(d) $2 e^{4}-e-a$
Show Answer
Solution: $\int _{\mathrm{e}}^{\mathrm{e}^{4}} \sqrt{\log _{\mathrm{e}} \mathrm{x}} \mathrm{dx}$; Put $\sqrt{\log _{\mathrm{e}} \mathrm{x}}=\mathrm{t}$
$\Rightarrow \log _{\mathrm{e}} \mathrm{x}=\mathrm{t}^{2} \Rightarrow \mathrm{x}=\mathrm{e}^{\mathrm{t}^{2}} \Rightarrow \mathrm{dx}=\mathrm{e}^{\mathrm{t}^{2}} 2 \mathrm{t} \mathrm{dt}$
$\int \limits _1^2 t\left(e^{t^2} 2 t\right) d t=\left(t \cdot e^{t^2}\right)_1^2-\int_1^2 1 e^{t^2} d t$
$=\left(2 \mathrm{e}^{4}-\mathrm{e}\right)-\int _{1}^{2} \mathrm{e}^{\mathrm{t}^{2}} \mathrm{dt}=2 \mathrm{e}^{4}-\mathrm{e}-\mathrm{a}$
Answer: d
4. $\int _{0}^{1} \dfrac{\sin t . d t}{t+1}=\alpha$, then the value of
$\int _{4 \pi-2}^{4 \pi} \dfrac{\sin \dfrac{\mathrm{t}}{2} \mathrm{dt}}{4 \pi+2-\mathrm{t}}$ is equal to
(a) $\alpha$
(b) $-\alpha$
(c) $\pi \alpha$
(d) $2 \alpha$
Show Answer
Solution: Put $4 \pi-t=2 z$ in the second integral
$=-\mathrm{dt}=2 \mathrm{dz}$
$\therefore-\int _{1}^{0} \dfrac{\sin (2 \pi-\mathrm{z}) 2 \mathrm{dz}}{2(\mathrm{z}+1)}=$
$=\int _{1}^{0} \dfrac{-\sin \mathrm{zdz}}{\mathrm{z}+1}=-\alpha$
Answer: b
5. Let $f$ be a continuous function. Let
$\begin{aligned} I_1=\int _{\sin ^2 t}^{1+\cos ^2 t}x f\{x(2-x)\} d x \text { and } I_2=\int _{\sin ^2 t}^{1+\cos ^2 t} f \{x(2-x)\} d x \text {, then } I_1: I_2= \end{aligned}$
(a) 0
(b) 1
(c) 2
(d) none
Show Answer
Solution: Apply $\int _{a}^{b} f(x) d x=\int _{a}^{b} f(a+b-x) d x$ in $I _{1}$
i.e: Replace $x$ by $1+\cos ^{2} t+\sin ^{2} t-x=2-x$
$\therefore \mathrm{I} _{1}=\int _{\sin ^{2} \mathrm{t}}^{1+\cos ^{2} \mathrm{t}}(2-\mathrm{x}) f\{(2-\mathrm{x})(2-(2-\mathrm{x}))\} \mathrm{dx}$
$.I _{1}=2 \int _{\sin ^{2} t}^{1+\cos ^{2} t} f(x(2-x)) d x-\int _{\sin ^{2} t}^{1+\cos ^{2} t} f(x(2-x)) d x$
$\Rightarrow \quad \mathrm{I} _{1}=2 \mathrm{I} _{2}-\mathrm{I} _{1}$ $2 \mathrm{I} _{1}=2 \mathrm{I} _{2} \Rightarrow \mathrm{I} _{1}=\mathrm{I} _{2}$
Answer: b
6. If $f(\mathrm{x})=\int \dfrac{\log _{\mathrm{e}} \mathrm{xdx}}{1+\mathrm{x}}$, then $f(\mathrm{e})+f\left(\dfrac{1}{\mathrm{e}}\right)$ is equal to
(a) 2
(b) $\dfrac{1}{2}$
(c) 1
(d) none
Show Answer
Solution: $f(x)+f\left(\dfrac{1}{\mathrm{x}}\right)=\int \dfrac{\log _{\mathrm{e}} \mathrm{x}}{1+\mathrm{x}} \mathrm{dx}+\int \dfrac{\log _{\mathrm{e}} \dfrac{1}{\mathrm{x}}}{1+\dfrac{1}{\mathrm{x}}} \mathrm{d}\left(\dfrac{1}{\mathrm{x}}\right)$
$=\int \dfrac{\log _{e} x}{1+x} d x+\int \dfrac{-\log _{e} x}{\dfrac{x+1}{x}}\left(-\dfrac{1}{x^{2}}\right) d x$
$\begin{aligned} & =\int \dfrac{\log _{e} \mathrm{x}}{1+\mathrm{x}} d x+\int \dfrac{\log _{e} \mathrm{x}}{\mathrm{x}(1+\mathrm{x})} \mathrm{dx} \\ \\ & =\int \dfrac{\log _{\mathrm{e}} \mathrm{x}}{1+\mathrm{x}}\left(1+\dfrac{1}{\mathrm{x}}\right) \mathrm{dx} \\ \\ & =\int \dfrac{\log _{\mathrm{e}} \mathrm{x}}{1+\mathrm{x}}\left(\dfrac{1+\mathrm{x}}{\mathrm{x}}\right) \mathrm{dx}=\dfrac{\left(\log _{\mathrm{e}} \mathrm{x}\right)^{2}}{2} \\ \\ & \therefore f(\mathrm{e})+f\left(\dfrac{1}{\mathrm{e}}\right)=\dfrac{\left(\log _{\mathrm{e}} \mathrm{e}\right)^{2}}{2}=\dfrac{1}{2} \end{aligned}$
Answer: b
7. $\int _{0}^{\mathrm{x}} f(\mathrm{t}) \mathrm{dt}=\mathrm{x}+\int _{\mathrm{x}}^{1} \mathrm{t} f(\mathrm{t}) \mathrm{dt}$, then $f(1)$ is equal to
(a) $\dfrac{1}{2}$
(b) 0
(c) 1
(d) $\dfrac{-1}{2}$
Show Answer
Solution: Differentiate both sides w.r.t $\mathrm{x}$
$\Rightarrow f(\mathrm{x})=1+(0-\mathrm{x} f(\mathrm{x})) \Rightarrow f(\mathrm{x})=\dfrac{1}{1+\mathrm{x}}$
$\therefore f(1)=\dfrac{1}{2}$
Answer: a
8. $f(\mathrm{x})=\cos \mathrm{x}-\int _{0}^{y}(\mathrm{x}-\mathrm{t}) f(\mathrm{t}) \mathrm{dt}$, then $f^{\prime \prime}(\mathrm{x})+f(\mathrm{x})$ is equal to
(a) $-\cos x$
(b) $-\sin x$
(c) 0
(d) none of these
Show Answer
Solution: $f(\mathrm{x})=\cos \mathrm{x}-\mathrm{x} \int _{0}^{\mathrm{x}} f(\mathrm{t}) \mathrm{dt}+\int _{0}^{\mathrm{x}} \mathrm{t} f(\mathrm{t}) \mathrm{dt}$
$\therefore f^{\prime}(\mathrm{x})$ = $-\sin \mathrm{x}-\left\{\mathrm{x}(f(\mathrm{x})-0)+\int _{0}^{\mathrm{x}}(\mathrm{t}) \mathrm{dt}\right\}+\{\mathrm{x} f(\mathrm{x})-0\}$
(applying lebnitz rule)
$\begin{aligned} & f^{\prime}(\mathrm{x})=-\sin \mathrm{x}-\mathrm{x} f(\mathrm{x})-\int _{0}^{\mathrm{x}} f(\mathrm{t}) \mathrm{dt}+\mathrm{x} f(\mathrm{x}) \\ & \therefore f^{\prime \prime}(\mathrm{x})=-\cos \mathrm{x}-(f(\mathrm{x}) 1-0) \\ & \Rightarrow f^{\prime \prime}(\mathrm{x})+f(\mathrm{x})=-\cos \mathrm{x} \end{aligned}$
Answer: a
Exercise:
1. Let $\mathrm{I}=\int _{0}^{1} \dfrac{\sin \mathrm{x}}{\sqrt{\mathrm{x}}} \mathrm{dx}$ and $\mathrm{J}=\int _{0}^{1} \dfrac{\cos \mathrm{x}}{\sqrt{\mathrm{x}}} \mathrm{dx}$. Then which one of the following is true?
(a) $\mathrm{I}<\dfrac{2}{3}$ & $\mathrm{~J}>2$
(b) $\mathrm{I}>\dfrac{2}{3}$ & $\mathrm{~J}<2$
(c) $\mathrm{I}>\dfrac{2}{3}$ & $\mathrm{~J}>2$
(d) $\mathrm{I}<\dfrac{2}{3}$ & $\mathrm{~J}<2$
Show Answer
Answer: d2. The equaiton of a curve is $\mathrm{y}=f(\mathrm{x})$. The tangent at $(1, f(1)),(2, f(2))$ & $(3, f(3))$ make angles $\dfrac{\pi}{6}, \dfrac{\pi}{3}$ & $\dfrac{\pi}{4}$ respectively with the positive direction of the $\mathrm{x}-$ axis. Then the value of $\int _{2}^{3} f^{\prime}(\mathrm{x}) f^{\prime \prime}(\mathrm{x}) \mathrm{dx}+\int _{1}^{3} f^{\prime \prime}(\mathrm{x}) \mathrm{dx}=$
(a) $\dfrac{1}{\sqrt{3}}$
(b) $\dfrac{-1}{\sqrt{3}}$
(c) 0
(d) none of these
Show Answer
Answer: b3. $\lim\limits _{\lambda \rightarrow 0}\left(\int _{0}^{1}(1+x)^{\lambda} d x\right)^{1 / \lambda}$ is equal to
(a) $2 \log _{\mathrm{e}} 2$
(b) $\dfrac{4}{\mathrm{e}}$
(c) $\log _{\mathrm{e}}(4 / \mathrm{e})$
(d) $4$
Show Answer
Answer: b4. If $\beta+2 \int _{0}^{1} x^{2} e^{-x^{2}} d x=\int _{0}^{1} e^{-x^{2}} d x$, then the value of $\beta$ is
(a) $e^{-1}$
(b) e
(c) $\dfrac{1}{2} \mathrm{e}$
(d) cannot be determined
Show Answer
Answer: a5.* $\quad$ If $f(a)=\int _{a-1}^{a+1} \dfrac{d x}{1+x^{8}}$, then the value of a for which $f(a)$ attains maximum is
(a) at $\mathrm{a}=0$
(b) at one value of a only
(c) at two values of a, one is $(-1,0)$ and the other is $(0,1)$
(d) at no value of a
Show Answer
Answer: a, b6.* If $I _{1}=\int _{0}^{\pi / 4}(\tan x)^{\cot x} d x, I _{2}=\int _{0}^{\pi / 4}(\cot x)^{\tan x} d x, I _{3}=\int _{0}^{\pi / 4}(\tan x)^{\tan x} d x, I _{4}=\int _{0}^{\pi / 4}(\cot x)^{\cot x} d x$ then
(a) $\mathrm{I} _{1}<\mathrm{I} _{3}$
(b) $\mathrm{I} _{2}<\mathrm{I} _{4}$
(c) $\mathrm{I} _{1}<\mathrm{I} _{4}$
(d) $\mathrm{I} _{3}<\mathrm{I} _{2}$
Show Answer
Answer: b, c7.* If $\mathrm{I}=\int _{0}^{1} \mathrm{e}^{\sin x} \mathrm{dx}$, then
(a) $\mathrm{I}>0$
(b) $\mathrm{I}<\mathrm{e}$
(c) $\quad\mathrm{I}>\mathrm{e}$
(d) $\mathrm{I}>\mathrm{e}^{2}$
Show Answer
Answer: a, b8. If $\int _{0}^{10} f(x) d x=5$, then $\sum\limits _{\mathrm{k}=1}^{10} \int _{0}^{1} f(\mathrm{k}-1+\mathrm{x}) \mathrm{d} \mathrm{x}=$
(a) 50
(b) 20
(c) 5
(d) 0
Show Answer
Answer: c9. The value of the definite integral $\dfrac{\int _{0}^{\pi}(\sqrt[2013]{\cos x}+\sqrt[2013]{\sin x}+\sqrt[2013]{\tan x}) d x}{\int _{0}^{\pi / 2} \sqrt[2013]{\sin x} d x}$ is
(a) 2
(b) 1
(c) $\dfrac{1}{2}$
(d) 0
Show Answer
Answer: a10. The value of $\int _{0}^{\pi} \dfrac{\sin \left(n+\dfrac{1}{2}\right) x}{\sin \dfrac{x}{2}} d x$ is
(a) $\dfrac{\pi}{2}$
(b) 0
(c) $\pi$
(d) $2 \pi$
Show Answer
Answer: c11. The value of $f(x)=\int _{0}^{\sin ^{2} x} \sin ^{-1} \sqrt{t} d t+\int _{0}^{\cos ^{2} x} \cos ^{-1} \sqrt{t} d t$ is
(a) $\dfrac{\pi}{2}$
(b) 1
(c) $\dfrac{\pi}{4}$
(d) none of these
Show Answer
Answer: c12. The value of the definite integral $\int _{0}^{1}\left(1+\mathrm{e}^{-\mathrm{x}^{2}}\right) \mathrm{dx}$ is
(a) -1
(b) 2
(c) $1+\mathrm{e}^{-1}$
(d) none of these