Integral Calculus - Definite Integrals (Lecture-02)
If $f(x)$ is a periodic function of period $T$; then
$\int \limits _{0}^{\mathrm{nT}} f(\mathrm{x}) \mathrm{dx}=\mathrm{n} \int \limits _{0}^{\mathrm{T}} f(\mathrm{x}) \mathrm{d} \mathrm{x}$
Solved Examples
1. $\int \limits _{0}^{10} \mathrm{e}^{\{x\}} \mathrm{dx}=$
(a) $10(\mathrm{e}-1)$
(b) 10
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Solution:
$\because f(\mathrm{x})$ is a periodic function of period 1 , then
$\int \limits _{0}^{1} \mathrm{e}^{\{x\}} \mathrm{dx} \quad = 10 \quad \int \limits _0^1 e^{x-[x]} d x$
$=10 \int \limits _{0}^{1} \mathrm{e}^{\mathrm{x}} \mathrm{dx} \quad = \quad 10\left(\mathrm{e}^{\mathrm{x}}\right) _{0}^{1}=10 .(\mathrm{e}-1)$
Answer: (a)
2. $\int \limits _{-\pi / 2}^{199 \pi / 2} \sqrt{1+\cos 2 \mathrm{x}} \mathrm{dx}$ is equal to
(a) $50 \sqrt{2}$
(b) $100 \sqrt{2}$
(c) $150 \sqrt{2}$
(d) $200 \sqrt{2}$
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Solution:
$\begin{aligned} \int \limits _{-\pi / 2}^{199 \pi / 2} \sqrt{1+\cos 2 x} d x= & \int \limits _{-\pi / 2}^{199 \pi / 2} \sqrt{2 \cos ^{2} x} d x=\sqrt{2} \int \limits _{\dfrac{-\pi}{2}}^{\dfrac{199 \pi}{2}}|\cos x| d x=\sqrt{2} \int \limits _{0}^{100 \pi}|\cos x| d x \\ \\ =100 \sqrt{2} \int \limits _{0}^{\pi}|\cos x| d x & =100 \sqrt{2} \hspace{0.2cm} 2 \int \limits _{0}^{\dfrac{\pi}{2}}|\cos x| d x \\ \\ = & 200 \sqrt{2}(\sin x) _{0}^{\dfrac{\pi}{2}}=200 \sqrt{2}\left(\sin \dfrac{\pi}{2}-\sin 0\right)=200 \sqrt{2} \end{aligned}$
Answer: (d)
3. Let $f$ be a real valued function satisfying $f(\mathrm{x})+f(\mathrm{x}+6)=f(\mathrm{x}+3)+f(\mathrm{x}+9)$. Then $\int \limits _{\mathrm{x}}^{\mathrm{x}+12} f(\mathrm{t}) \mathrm{dt}$ is
(a) linear function of $\mathrm{x}$
(b) exponential function of $\mathrm{x}$
(c) a constant
(d) None of these
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Solution:
$\because f(\mathrm{x})+f(\mathrm{x}+6)=f(\mathrm{x}+3)+f(\mathrm{x}+9)$ …………………(1)
Replace $x$ by $x+3$
$f(\mathrm{x}+3)+f(\mathrm{x}+9)=f(\mathrm{x}+6)+f(\mathrm{x}+12)$. …………………(2)
Adding (1) and (2) we get
$f(\mathrm{x})=f(\mathrm{x}+12)$
$\Rightarrow f(\mathrm{x})$ is a periodic function of period 12
Let $\mathrm{g}(\mathrm{x})=\int \limits _{\mathrm{x}}^{\mathrm{x}+12} f(\mathrm{t}) \mathrm{dt}$
$\therefore \mathrm{g}^{1}(\mathrm{x})=f(\mathrm{x}+12)-f(\mathrm{x})=0(\because f(\mathrm{x})=f(\mathrm{x}+12))$
$\Rightarrow \mathrm{g}(\mathrm{x})$ is constant
Answer: (c)
4. The value of $\int \limits _{-\pi / 4}^{n \pi-\pi / 4}|\sin x+\cos x| d x$ is
(a) $\sqrt{2} \mathrm{n}$
(B) $2 \sqrt{2} \mathrm{n}$
(c) $4 \sqrt{2} \mathrm{n}$
(d) None of these
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Solution:
$\int \limits _{-\pi / 4}^{n \pi-\pi / 4} \sqrt{2}\left|\sin \left(x+\dfrac{\pi}{4}\right)\right| d x$
Put $\mathrm{x}+\dfrac{\pi}{4}=\mathrm{t}$
$\Rightarrow \sqrt{2} \int \limits _{0}^{\mathrm{nn}}|\operatorname{sint}| \mathrm{dt}=\sqrt{2} \mathrm{n} \int \limits _{0}^{\pi}|\operatorname{sint}| \mathrm{dt}$
$=\sqrt{2} \mathrm{n}(-\operatorname{cost}) _{0} \pi=\sqrt{2} \mathrm{n}(-\cos \pi+\cos 0)=2 \sqrt{2} \mathrm{n}$.
Answer: (b)
5. If $\int \limits _{a}^{b}|\sin x| d x=8$ and $\int \limits _{0}^{a+b}|\cos x| d x=9$
then the value of $a$ and $b$ are
(a) $\dfrac{\pi}{4}, \dfrac{17 \pi}{4}$
(b) $\dfrac{3 \pi}{4}, \dfrac{19 \pi}{4}$
(c) $\dfrac{\pi}{6}, \dfrac{25 \pi}{6}$
(d) None of these
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Solution:
$\mathrm{b}-\mathrm{a}=4 \pi$ & $(\mathrm{a}+\mathrm{b})-0=\dfrac{9 \pi}{2}$
Solving we get $\quad \quad \mathrm{a}=\dfrac{\pi}{4} \quad \quad \mathrm{~b}=\dfrac{17 \pi}{4}$
6. If $\mathrm{A}=\int \limits _{1}^{\sin \theta} \dfrac{\mathrm{tdt}}{1+\mathrm{t}^{2}}$ & $\mathrm{~B}=\int \limits _{1}^{\operatorname{cosec} \theta} \dfrac{\mathrm{dt}}{\mathrm{t}\left(1+\mathrm{t}^{2}\right)}$, then the value of $\left|\begin{array}{ccc}\mathrm{A} & \mathrm{A}^{2} & \mathrm{~B} \\ \mathrm{e}^{\mathrm{A}} \mathrm{e}^{\mathrm{B}} & \mathrm{B}^{2} & -1 \\ 1 & \mathrm{~A}^{2}+\mathrm{B}^{2} & -1\end{array}\right|$ is
(a) $\sin \theta$
(b) $\operatorname{cosec} \theta$
(c) 0
(d) 1
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Solution:
Put $\mathrm{t}=\dfrac{1}{\mathrm{z}}$ in $\mathrm{B} \Rightarrow \mathrm{dt}=\dfrac{-1}{\mathrm{z}^{2}} \mathrm{dz}$
$\therefore$ the integral $\mathrm{B}$ reduces to
$\int \limits _{1}^{\sin \theta} \dfrac{-\dfrac{1}{\mathrm{z}^{2}} \mathrm{dz}}{\dfrac{1}{\mathrm{z}}\left(1+\dfrac{1}{\mathrm{z}^{2}}\right)}=-\int \limits _{1}^{\sin \theta} \dfrac{\mathrm{zdz}}{1+\mathrm{z}^{2}}=-\int \limits _{1}^{\sin \theta} \dfrac{\mathrm{tdt}}{1+\mathrm{t}^{2}}=-\mathrm{A}$
$\therefore \mathrm{A}+\mathrm{B}=0$
$\left|\begin{array}{ccc}\mathrm{A} & \mathrm{A}^{2} & \mathrm{~B} \\ \mathrm{e}^{\mathrm{A}} \mathrm{e}^{\mathrm{B}} & \mathrm{B}^{2} & -1 \\ 1 & \mathrm{~A}^{2}+\mathrm{B}^{2} & -1\end{array}\right|=\left|\begin{array}{ccc}\mathrm{A} & \mathrm{A}^{2} & -\mathrm{A} \\ 1 & \mathrm{~A}^{2} & -1 \\ 1 & 2 \mathrm{~A}^{2} & -1\end{array}\right|=0$
(Applying $\mathrm{C} _{1} \rightarrow \mathrm{C} _{1}+\mathrm{C} _{3}$ )
7. $\int \limits _{e^{-1}}^{\tan x} \dfrac{\mathrm{t} d \mathrm{t}}{1+\mathrm{t}^{2}}+\int \limits _{\mathrm{e}^{-1}}^{\cos \mathrm{x}} \dfrac{\mathrm{dt}}{\mathrm{t}\left(1+\mathrm{t}^{2}\right)}=$
(a) 1
(b) e
(c) $\mathrm{e}^{-1}$
(d) $\dfrac{\pi}{4}$
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Solution:
Put $\mathrm{t}=\dfrac{1}{\mathrm{z}}$ in 2nd integral
$\therefore \mathrm{dt}=-\dfrac{1}{\mathrm{z}^{2}} \mathrm{dz}$
2nd integral reduces to $\int \limits _{\mathrm{e}}^{\tan x} \dfrac{\dfrac{-1}{\mathrm{z}^{2}} \mathrm{dz}}{\dfrac{1}{\mathrm{z}}\left(1+\dfrac{1}{\mathrm{z}^{2}}\right)}=\int \limits _{\operatorname{tanx}}^{e} \dfrac{\mathrm{z} \cdot \mathrm{dz}}{1+\mathrm{z}^{2}}$
$\therefore \int \limits _{\dfrac{1}{e}}^{\tan x} \dfrac{\mathrm{tdt}}{1+\mathrm{t}^{2}}+\int \limits _{\tan \mathrm{x}}^{\mathrm{e}} \dfrac{\mathrm{tdt}}{1+\mathrm{t}^{2}}=\int \limits _{\dfrac{1}{\mathrm{c}}}^{\mathrm{e}} \dfrac{\mathrm{t} \cdot \mathrm{dt}}{1+\mathrm{t}^{2}}$
$=\dfrac{1}{2}\left(\log _{\mathrm{e}}\left(1+\mathrm{t}^{2}\right)\right) _{\dfrac{1}{\mathrm{e}}}{ }^{\mathrm{e}}=\dfrac{1}{2} \log \mathrm{e}\left(\dfrac{1+\mathrm{e}^{2}}{1+\mathrm{e}^{-2}}\right)=\dfrac{1}{2} \log _{\mathrm{e}} \mathrm{e}^{2}=1$
Answer : (a)
Exercise
1. $\mathrm{I}=\int \limits _{0}^{10} \sin {\mathrm{x}} \mathrm{dx}$
(a) 10
(b) $10(1-\cos 1)$
(c) $5(1-\cos 1)$
(d) $5 \cos 1$
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Answer: b2. $\int \limits _{\pi}^{10 \pi}|\sin x| d x=$
(a) 20
(b) 8
(c) 10
(d) 18
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Answer: d3. If $\mathrm{I} _{1}=\int \limits _{0}^{3 \pi} f\left(\cos ^{2} \mathrm{x}\right) \mathrm{dx}$ & $\mathrm{I} _{2}=\int \limits _{0}^{\pi} f\left(\cos ^{2} \mathrm{x}\right) \mathrm{dx}$ then
(a) $\mathrm{I} _{1}=\mathrm{I} _{2}$
(b) $\mathrm{I} _{1}=2 \mathrm{I} _{2}$
(c) $\mathrm{I} _{1}=5 \mathrm{I} _{2}$
(d) $\mathrm{I} _{1}=3 \mathrm{I} _{2}$
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Answer: d4. $\quad \lim \limits _{n \rightarrow \infty} \dfrac{(n !)^{\dfrac{1}{n}}}{n^{n}}$ is equal to
(a) $\mathrm{e}$
(b) $\dfrac{1}{\mathrm{e}}$
(c) e-1
(d) $\mathrm{e}-5$
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Answer: b5. The value of the definite integral $\int \limits _{0}^{1} \dfrac{\mathrm{xdx}}{\mathrm{x}^{3}+16}$ lies in the interval $[\mathrm{a}, \mathrm{b}]$. Then smallest such interval is
(a) $\left[0, \dfrac{1}{17}\right]$
(b) $[0,1]$
(c) $\left[0, \dfrac{1}{27}\right]$
(d) $\left[\dfrac{1}{17}, 1\right]$
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Answer: a6. Let $f: \mathrm{R} \rightarrow \mathrm{R}$ such that $f(\mathrm{x}+2 \mathrm{y})=f(\mathrm{x})+f(2 \mathrm{y})+4 \mathrm{xy}, \forall \mathrm{x}, \mathrm{y} \in \mathrm{R}$ & $f(0)=0$ If $\mathrm{I} _{1}=\int \limits _{0}^{1} f(\mathrm{x}) \mathrm{dx}$,
$\mathrm{I} _{2}=\int \limits _{0}^{1} f(\mathrm{x}) \mathrm{dx}$ & $\mathrm{I} _{3}=\int \limits _{\dfrac{1}{2}}^{2} f(\mathrm{x}) \mathrm{dx}$, then
(a) $\mathrm{I} _{1}=\mathrm{I} _{2}>\mathrm{I} _{3}$
(b) $\mathrm{I} _{1}>\mathrm{I} _{2}>\mathrm{I} _{3}$
(c) $\mathrm{I} _{1}=$ $\mathrm{I} _{2}<\mathrm{I} _{3}$
(d) $\mathrm{I} _{1}<$ $\mathrm{I} _{2}<$ $\mathrm{I} _{3}$
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Answer: c7. The integral $\int \limits _{\tan ^{-1} \lambda}^{\cot ^{-1} \lambda} \dfrac{\tan x}{\tan x+\cot x} d x, \forall \lambda \in \mathrm{R}$ can not take the value
(a) $\dfrac{-\pi}{4}$
(b) $\dfrac{-\pi}{2}$
(c) $\dfrac{\pi}{4}$
(d) $\dfrac{3 \pi}{4}$
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Answer: a, b, d8. Read the following and answer the questions that follow :-
$I=\int \limits _{0}^{10 \pi} \dfrac{\cos 6 x \cos 7 x \cos 8 x \cos 9 x}{1+e^{2 \sin ^{3} 4 x}} d x$
(i) If $I=k \int \ limits _{0}^{\dfrac{\pi}{2}} \cos 6 \mathrm{x} \cos 7 \mathrm{x} \cos 8 \mathrm{x} \cos 9 \mathrm{xdx}$, then $\mathrm{k}=$
(a) 5
(b) 10
(c) 20
(d) None
(ii) If $I=c \int \limits _{0}^{\dfrac{\pi}{2}} \cos 6 x \cos 8 x \cos 2 x d x$, then $c=$
(a) 5
(b) 10
(c) 20
(d) None
(iii) The value of $\mathrm{I}=$
(a) $\dfrac{5 \pi}{4}$
(b) $\dfrac{5 \pi}{8}$
(c) $\dfrac{5 \pi}{16}$
(d) $\dfrac{5 \pi}{12}$
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Answer: (i) b (ii) a (iii) b9. Match the following if
$I _{1}=\int \limits _{0}^{1} \dfrac{e^{\tan ^{-1} x}}{\sqrt{1+\mathrm{x}^{2}}} d x, \quad I _{2}=\int \limits _{0}^{1} \dfrac{x^{\tan ^{-1} x}}{\sqrt{1+\mathrm{x}^{2}}} d x, \quad I _{3}=\int \limits _{0}^{1} \dfrac{x^{2} e^{\tan ^{-1} x}}{\sqrt{1+\mathrm{x}^{2}}} d x$, then
| Column I | Column II | ||
|---|---|---|---|
| (a) | $\mathrm{I} _{1}+\mathrm{I} _{2}$ | (p) | $3 / 2$ |
| (b) | $\mathrm{I} _{1}+\mathrm{I} _{2}-4 \mathrm{I} _{3}$ | (q) | 1 |
| (c) | $\dfrac{\mathrm{I} _{1}+\mathrm{I} _{2}+1}{\mathrm{I} _{1}+\mathrm{I} _{2}+2 \mathrm{I} _{3}}$ | (r) | $\sqrt{2} \mathrm{e}^{\pi / 4}-1$ |
| (d) | $\dfrac{2\left(\mathrm{I} _{1}+\mathrm{I} _{2}\right)}{2 \mathrm{I} _{3}+1}$ | (s) | $\sqrt{2} \mathrm{e}^{-\pi / 4}$ |
| (t) | $\sqrt{2} \mathrm{e}^{\pi / 4}-3$ |
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Answer: $\mathrm{a} \rightarrow \mathrm{r}, \mathrm{b} \rightarrow \mathrm{t}, \mathrm{c} \rightarrow \mathrm{q}, \mathrm{d} \rightarrow \mathrm{r}$10. Let $S _{n}=\sum \limits _{k=0}^{n} \dfrac{n}{n^{2}+k n+k^{2}}$ & $T _{n}=\sum \limits _{k=0}^{n-1} \dfrac{n}{n^{2}+k n+k^{2}}$ for $n=1,2,3$ then
(a) $\mathrm{S} _{\mathrm{n}}<\dfrac{\pi}{3 \sqrt{3}}$
(b) $\mathrm{S} _{\mathrm{n}}>\dfrac{\pi}{3 \sqrt{3}}$
(c) $\mathrm{T} _{\mathrm{n}}<\dfrac{\pi}{3 \sqrt{3}}$
(d) $\mathrm{T} _{\mathrm{n}}>\dfrac{\pi}{3 \sqrt{3}}$
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Answer: a, d11. $\int \limits _{0}^{[x]} \dfrac{2^{x}}{2^{[x]}} d x$ is
(a) $[\mathrm{x}] \log _{\mathrm{e}} 2$
(b) $\dfrac{[\mathrm{x}]}{\log _{\mathrm{e}} 2}$
(c) $\dfrac{1}{2} \dfrac{[\mathrm{x}]}{\log _{\mathrm{e}} 2}$
(d) None of these.
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Answer: b12. The number of positive continuous functions $f(\mathrm{x})$ defined in $[0,1]$ for which $\int \limits _{0}^{1} f(\mathrm{x}) \mathrm{dx}=1$,
$\int \limits _{0}^{1} \mathrm{x} f(\mathrm{x}) \mathrm{dx}=\mathrm{a}, \int \limits _{0}^{1} \mathrm{x}^{2} f(\mathrm{x}) \mathrm{d} \mathrm{x}=\mathrm{a}^{2}$ is
(a) one
(b) infinite
(c) two
(d) zero
BETA
