Integral Calculus Definite Integrals(Lecture-03)
Property:
Limit as a sum
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Express the given series in the form of $\sum \dfrac{1}{\mathrm{n}} f\left(\dfrac{\mathrm{r}}{\mathrm{n}}\right)$
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Then the limit as its sum when $\mathrm{n} \rightarrow \infty$ i.e. $\lim \limits _{\mathrm{n} \rightarrow \infty} \sum \dfrac{1}{\mathrm{n}} . f\left(\dfrac{\mathrm{r}}{\mathrm{n}}\right)$
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Replace $\dfrac{\mathrm{r}}{\mathrm{n}}$ by $x$ and $\dfrac{1}{\mathrm{n}}$ by $\mathrm{dx}$ and $\dfrac{1}{\mathrm{n}}$ by the sign of $\int$
Solved Examples
1. $\lim \limits _{\mathrm{n} \rightarrow \infty} \dfrac{1}{\mathrm{n}+1}+\dfrac{1}{\mathrm{n}+2}+\dfrac{1}{\mathrm{n}+3}+—-+\dfrac{1}{2 \mathrm{n}}=$
(a) 1
(b) 2
(c) $\log _{\mathrm{e}} 2$
(d) $\log _{e} 4$
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Solution: $\lim \limits _{n \rightarrow \infty} \sum \limits _{r=1}^{n} \dfrac{1}{n+r}=\lim \limits _{n \rightarrow \infty} \dfrac{1}{n} \sum \limits _{r=1}^{n} \dfrac{1}{1+\dfrac{r}{n}}$
$=\int _{0}^{1} \dfrac{d x}{1+x}=\log _{\mathrm{e}} 2$
Answer: c
2. The value of $\sum \limits _{\mathrm{r}=0}^{\mathrm{n}-1} \dfrac{1}{\sqrt{4 \mathrm{n}^{2}-\mathrm{r}^{2}}}$ as $\mathrm{n} \rightarrow \infty$ is
(a) $\dfrac{\pi}{3}$
(b) $\dfrac{\pi}{6}$
(c) $\dfrac{\pi}{4}$
(d) none of these
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Solution: $\sum \limits _{\mathrm{r}=0}^{\mathrm{n}-1} \dfrac{1}{\sqrt{4 \mathrm{n}^{2}-\mathrm{r}^{2}}}=\dfrac{1}{\mathrm{n}} \sum \limits _{\mathrm{r}=0}^{\mathrm{n}-1} \dfrac{1}{\sqrt{4-\left(\dfrac{\mathrm{r}}{\mathrm{n}}\right)^{2}}}=\int _{0}^{1} \dfrac{\mathrm{dx}}{\sqrt{4-\mathrm{x}^{2}}}=\left(\sin ^{-1} \dfrac{\mathrm{x}}{2}\right) _{0}^{1}$
$=\sin ^{-1} \dfrac{1}{2}-\sin ^{-1} 0=\dfrac{\pi}{6}$
Answer: b
3. The value of $\lim \limits _{\mathrm{n} \rightarrow \infty} \dfrac{\left(1^{2}+2^{2}+3^{2}+-+\mathrm{n}^{2}\right)\left(1^{3}+2^{3}+-+\mathrm{n}^{3}\right)}{1^{6}+2^{6}+—–+\mathrm{n}^{6}}$
(a) 1
(b) $\dfrac{7}{12}$
(c) $\dfrac{5}{6}$
(d) none
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Solution: $\lim \limits _{n \rightarrow \infty} \dfrac{\left(\sum \limits _{r=1}^n r^2 \cdot \sum \limits _{r=1}^n r^3\right)}{\sum \limits _{r=1}^n r^6}$
$=\lim \limits _{n \rightarrow \infty} \dfrac{\left(\dfrac{1}{n} \sum \limits _{r=1}^n\left(\dfrac{r}{n}\right)^2\right)\left(\dfrac{1}{n} \sum \limits _{r=1}^n\left(\dfrac{r}{n}\right)^3\right)}{\dfrac{1}{n} \sum \limits _{r=1}^n\left(\dfrac{r}{n}\right)^6}$
$=\dfrac{\int_0^1 x^2 d x \int_0^1 x^3 d x}{\int_0^1 x^6 d x}=\dfrac{7}{12}$
4. $\int _{0}^{1}\left[\dfrac{2}{\mathrm{e}^{\mathrm{x}}}\right] \mathrm{dx}$ is equal to
(a) $\log _{e} 2$
(b) $\mathrm{e}^{2}$
(c) 0
(d) $\dfrac{2}{\mathrm{e}}$
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Solution: $\int _{0}^{2}\left[2 \mathrm{e}^{-\mathrm{x}}\right] \mathrm{dx}=\int _{0}^{\log _{c} 2} 1 \mathrm{dx}+\int _{\log _{c} 2}^{\infty} 0 \mathrm{dx}$
$=\log _{\mathrm{e}} 2$
Answer: a
5. The value of $\int _{0}^{\pi / 3}[\sqrt{3} \tan x] d x$ is
(a) $\dfrac{5 \pi}{6}$
(b) $\dfrac{5 \pi}{6}-\tan ^{-1} \dfrac{2}{\sqrt{3}}$
(c) $\dfrac{\pi}{2}-\tan ^{-1} \dfrac{2}{\sqrt{3}}$
(d) none
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Solution: $\int _{0}^{\pi / 3}[\sqrt{3} \tan x] d x=\int _{0}^{\pi / 6} 0 \mathrm{dx}+\int _{\pi / 6}^{\tan ^{-1} \dfrac{2}{3}} 1 \mathrm{dx}+\int _{\tan ^{-1} \dfrac{2}{\sqrt{3}}}^{\pi / 3} 2 \mathrm{dx}$
$\begin{aligned} & =0+\left(\tan ^{-1} \dfrac{2}{\sqrt{3}}-\dfrac{\pi}{6}\right)+2\left(\dfrac{\pi}{3}-\tan ^{-1} \dfrac{2}{\sqrt{3}}\right) \\ & =\quad \dfrac{\pi}{2}-\tan ^{-1} \dfrac{2}{\sqrt{3}} \end{aligned}$
Answer: c
6. $\int _{-1}^{1}[\mathrm{x}[1+\sin \pi \mathrm{x}]+1] \mathrm{dx}$ is equal to
(a) 2
(b) 0
(c) 1
(d) none
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Solution: $\int _{-1}^{0} 1 \cdot \mathrm{dx}+\int _{0}^{1}[\mathrm{x}+1] \cdot \mathrm{dx}$
$=\int _{-1}^{0}[x \cdot 0+1] d x+\int _{0}^{1}[x \cdot 1+1] d x$
$\begin{array}{ll} = & \int _{-1}^{0} 1 \mathrm{dx}+\int _{0}^{1}([\mathrm{x}]+1) \mathrm{dx} \\ \\ = & (\mathrm{x}) _{-1}^{0}+(\mathrm{x}) _{0}^{1} \\ \\ = & 1+1=2 \end{array}$
Answer: a
7. $\int _{0}^{2}\left[\mathrm{x}^{2}-1\right] \mathrm{dx}$ is equal to
(a) $3-\sqrt{3}-\sqrt{2}$
(b) 2
(c) 1
(d) $3-2 \sqrt{3}-3 \sqrt{2}$
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Solution: $\int _{0}^{1}(-1) \mathrm{dx}+\int _{1}^{\sqrt{2}} 0 \mathrm{dx}+\int _{\sqrt{2}}^{\sqrt{3}} 1 \mathrm{dx}+\int _{\sqrt{3}}^{2} 2 \mathrm{dx}$
$\begin{aligned} & =-(1-0)+0+1(\sqrt{3}-\sqrt{2})+2(2-\sqrt{3}) \\ \\ & =-1+\sqrt{3}-\sqrt{2}+4-2 \sqrt{3} \\ \\ & = 3-\sqrt{3}-\sqrt{2} \end{aligned}$
Answer: a
Exercise
1. $\lim \limits _{\mathrm{n} \rightarrow \infty} \dfrac{1^{\mathrm{P}}+2^{\mathrm{P}}+3^{\mathrm{P}}+……+\mathrm{n}^{\mathrm{P}}}{\mathrm{n}^{\mathrm{P}}+1}=$
(a) $\dfrac{1}{\mathrm{P}+1}$
(b) $\dfrac{1}{1-\mathrm{P}}$
(c) $\dfrac{1}{\mathrm{P}}-\dfrac{1}{\mathrm{P}-1}$
(d) $\dfrac{\mathrm{P}}{\mathrm{P}+1}$
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Answer: a2. Let a,b,c be non-zero real numbers such that $\int _{0}^{1}\left(1+\cos ^{8} \mathrm{x}\right)\left(a x^{2}+\mathrm{bx}+\mathrm{c}\right) \mathrm{dx}$
$=\int _{0}^{2}\left(1+\cos ^{8} \mathrm{x}\right)\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right) \mathrm{dx}$. Then the quadratic equation $\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0$ has
(a) no root in $(0,2)$
(b) at least one root in $(1,2)$
(c) a double root in $(0,2)$
(d) two imaginary roots
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Answer: b3. For every function $f(\mathrm{x})$ which is twice differentiable these will be good approximation of
$\begin{aligned} & \int _{\mathrm{a}}^{\mathrm{b}} f(\mathrm{x}) \mathrm{dx}=\dfrac{\mathrm{b}-\mathrm{a}}{2}(f(\mathrm{a})+f(\mathrm{~b})), \text { for more accurate results for } \mathrm{c} \in(\mathrm{a}, \mathrm{b}) \\ & \mathrm{F}(\mathrm{c})=\dfrac{\mathrm{c}-\mathrm{a}}{2}(f(\mathrm{a})-f(\mathrm{c}))+\dfrac{\mathrm{b}-\mathrm{c}}{2}(f(\mathrm{~b})-f(\mathrm{c})) \quad \text { where } \quad \mathrm{c}=\dfrac{\mathrm{a}+\mathrm{b}}{2} \\ & \int _{\mathrm{a}}^{\mathrm{b}} f(\mathrm{x}) \mathrm{dx}=\dfrac{\mathrm{b}-\mathrm{a}}{4}(f(\mathrm{a})+f(\mathrm{~b})+2 f(\mathrm{c})) \mathrm{dx} \end{aligned}$
i. Good approximation of $\int _{0}^{\pi / 2} \sin x d x$ is
(a) $\dfrac{\pi}{4}$
(b) $\dfrac{\pi(\sqrt{2}+1)}{4}$
(c) $\dfrac{\pi(\sqrt{2}+1)}{8}$
(d) $\dfrac{\pi}{8}$
ii. If $f^{\prime \prime}(\mathrm{x})<0, \forall \mathrm{x} \in(\mathrm{a}, \mathrm{b})$ & $ (\mathrm{c}, f(\mathrm{c}))$ is point of maximum where $\mathrm{c} \in(\mathrm{a} . \mathrm{b})$ then $f^{\prime}(\mathrm{c})$ is
(a) $\dfrac{f(\mathrm{~b})-f(\mathrm{a})}{\mathrm{b}-\mathrm{a}}$
(b) $3\left(\dfrac{f(\mathrm{~b})-f(\mathrm{a})}{\mathrm{b}-\mathrm{a}}\right)$
(c) $2\left(\dfrac{f(\mathrm{~b})-f(\mathrm{a})}{\mathrm{b}-\mathrm{a}}\right)$
(d) 0
iii. If $\lim \limits _{\mathrm{t} \rightarrow \mathrm{a}} \dfrac{\int \limits _{\mathrm{a}}^{\mathrm{t}} f(\mathrm{x}) \mathrm{dx}-\dfrac{\mathrm{t}-\mathrm{a}}{2}(f(\mathrm{t})+f(\mathrm{a}))}{(\mathrm{t}-\mathrm{a})^{3}}=0$, then degree of polynomial function $f(\mathrm{x})$ at most is
(a) 0
(b) 1
(c) 3
(d) 2
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Answer: (i) c (ii) a (iii) b4. Match the following:-
Column I | Column II | ||
---|---|---|---|
a. | $\int _{0}^{\infty} \mathrm{e}^{-4 \mathrm{x}} \sin 5 \mathrm{x} d \mathrm{x}$ | (p) | 3 |
b. | $\int _{2}^{8} \dfrac{\left[x^{2}\right] d x}{\left[x^{2}-20 x+100\right]+\left[x^{2}\right]}$ | (q) | $\dfrac{5}{41}$ |
c. | $\int _{0}^{3 \pi / 2}|\sin x| d x, n \in N$ | (r) | 120 |
d. | $\int _{0}^{\infty} x^{5} e^{-x} d x$ | (s) | 60 |
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Answer: $\mathrm{a} \rightarrow \mathrm{q} ; \mathrm{b} \rightarrow \mathrm{p} ; \mathrm{c} \rightarrow \mathrm{p} ; \mathrm{d} \rightarrow \mathrm{r} ;$5. $\int _{1}^{3} \dfrac{d x}{x^{2}+[x]^{2}+1-2 x[x]}=$
(a) $\pi$
(b) $3 \pi$
(c) $8 \pi$
(d) $\dfrac{\pi}{2}$
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Answer: d6.* If $I _{n}=\int _{-\pi}^{\pi} \dfrac{\sin n x}{\left(1+\pi^{x}\right) \sin x} d x, n=0,1,2, \ldots \ldots$ then
(a) $I _{n}=I _{n+2}$
(b) $\sum \limits _{\mathrm{m}=1}^{10} \mathrm{I} _{2 \mathrm{~m}+1}=10 \pi$
(c) $\sum \limits _{\mathrm{m}=1}^{10} \mathrm{I} _{2 \mathrm{~m}}=0$
(d) $I _{n}=I _{n+1}$
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Answer: a, b, c7. All the values of a for which $\int _{1}^{2}\left(a^{2}+(4-4 a) x+4 x^{3}\right) d x \leq 12$ are given by
(a) $\mathrm{a} \leq 4$
(b) $a=3$
(c) $0 \leq \mathrm{a}<3$
(d) none of these
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Answer: b8. Value of $\int _{0}^{2}\left[x^{2}-x+1\right] d x$ is
(a) $\dfrac{7+\sqrt{5}}{2}$
(b) $\dfrac{5-\sqrt{5}}{2}$
(c) $\dfrac{7-\sqrt{5}}{2}$
(d) none of these
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Answer: b9. If $I=\int _{0}^{1} \dfrac{d x}{\left(5+2 x-2 x^{2}\right)\left(1+e^{2-4 x}\right)}$ and $I _{1}=\int _{0}^{1} \dfrac{d x}{5+2 x-2 x^{2}} d x$ then
(a) $\mathrm{I}=\dfrac{1}{2} \mathrm{I} _{1}$
(b) $\mathrm{I}=2 \mathrm{I} _{1}$
(c) $3 \mathrm{I}=5 \mathrm{I} _{1}$
(d) none of these
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Answer: a10. Match the following:-
Column I | Column II | ||
---|---|---|---|
a. | $\lim \limits _{n \rightarrow \infty} \sum \limits _{\mathrm{r}=1}^{\mathrm{n}} \dfrac{\mathrm{r}^{2}}{\mathrm{r}^{3}+\mathrm{n}^{3}}$ | (p) | $\log _{e} 2$ |
b. | $\lim \limits _{\mathrm{n} \rightarrow \infty} \dfrac{1}{\mathrm{n}}\left(\tan \dfrac{\pi}{4 \mathrm{n}}+\tan \dfrac{2 \pi}{4 \mathrm{n}}+\ldots . .+\tan \dfrac{\pi}{4}\right)$ | (q) | $2 \log _{\mathrm{e}} 2$ |
c. | $\lim \limits _{\mathrm{n} \rightarrow \infty} \dfrac{1}{\mathrm{n}}\left(1+\dfrac{1}{\mathrm{n}^{2}}\right)^{2 / \mathrm{n}^{2}}\left(1+\dfrac{2^{2}}{\mathrm{n}^{2}}\right)^{4 / \mathrm{n}^{2}} \ldots . .\left(1+\dfrac{\mathrm{n}^{2}}{\mathrm{n}^{2}}\right)^{2 \mathrm{n} / \mathrm{n}^{2}}$ | (r) | $\dfrac{2}{\pi} \log _{\mathrm{e}} 2$ |
d. | $\int _{0}^{\infty}\left[2 \mathrm{e}^{-\mathrm{x}}\right] \mathrm{dx}$ | (s) | $\dfrac{1}{3} \log _{\mathrm{e}} 2$ |
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Answer: $\mathrm{a} \rightarrow \mathrm{s} ; \mathrm{b} \rightarrow \mathrm{r} ; \mathrm{c} \rightarrow \mathrm{q} ; \mathrm{d} \rightarrow \mathrm{p} ;$11. $\int _{0}^{2}\left[\tan ^{-1} x\right] d x+\int _{0}^{2}\left[\cot ^{-1} x\right] d x=$
(a) $1-\cot 2$
(b) $1+\cot 2$
(c) $2(1+\cot 2)$
(d) $2(1-\cot 2)$
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Answer: c12.* Match the following
([.] denotes the greatest integer function)
Column I | Column II | ||
---|---|---|---|
a. | If $\mu<\int _{0}^{1} \dfrac{x^{7} d x}{\sqrt[3]{1+x^{8}}}<\lambda$, then | (p) | $[\lambda+\mu]=2$ |
b. | If $\mu<\int _{0}^{1} \dfrac{d x}{\sqrt{1+x^{6}}}<\lambda$, then | (q) | $[\lambda+\mu]=4$ |
c. | If $\mu<\int _{0}^{1} \dfrac{d x}{\sqrt{4-x^{2}-x^{3}}}<\lambda$, then | (r) | $[\lambda-\mu]=0$ |
(s) | $[\lambda-\mu]=3$ | ||
(t) | $[\lambda+\mu]=0$ |