Mathematics

# Evaluate the following integral as a limit of sum:$\displaystyle \int_{0}^{1} \ (x+x^2) dx$

##### SOLUTION
Let $I=\displaystyle\int^{1}_0 (x+x^2) d x$

As we know that

$\displaystyle\int ^{b}_a f(x) d x=(b-a)\lim_{n\to \infty} \dfrac{1}{n}(f(a)+f(a+h)+f(a+2 h)+\cdots+f(a+(n-1)h))$   where $h=\dfrac{b-a}{n}$

Here $a=0,b=1,h=\dfrac{b-a}{n}=\dfrac{1-0}{n}=\dfrac{1}{n}$ and $f(x)=x^2+x$

So $I=(1-0)\displaystyle\lim_{n\to \infty} \dfrac{1}{n}(f(0)+f(h)+(f(2 h)+\cdots+f((n-1)h))$

$f(0)=0^2+0=0$

$f(h)=h^2+h=\dfrac{1}{n^2}+\dfrac{1}{n}$

$f((n-1)h)=(n-1)^2 h^2+(n-1)h=\dfrac{(n-1)^2}{n^2}+\dfrac{n-1}{n}$

$\therefore I=\displaystyle\lim_{n\to \infty}\dfrac{1}{n}\bigg(0+\dfrac{1}{n^2}+\dfrac{1}{n}+\dfrac{4}{n^2}+\dfrac{2}{n}+\cdots+\dfrac{(n-1)^2}{n^2}+\dfrac{n-1}{n}\bigg)$

$=\displaystyle\lim_{n\to \infty}\dfrac{1}{n}\bigg(\bigg(\dfrac{1}{n^2}+\dfrac{2^2}{n^2}+\dfrac{3^2}{n^2}+\cdots+\dfrac{(n-1)^2}{n^2}\bigg)+\bigg(\dfrac{1}{n}+\dfrac{2}{n}+\dfrac{3}{n}+\cdots+\dfrac{n-1}{n}\bigg)\bigg)$

$=\displaystyle\lim_{n\to \infty}\dfrac{1}{n}\bigg(\dfrac{1}{n^2}(1^2+2^2+3^2+\cdots+(n-1)^2+\dfrac{1}{n}(1+2+3+\cdots+(n-1))\bigg)$

As we know that

$1+2+3+\cdots+n=\dfrac{n(n+1)}{2}$ and $1^2+2^2+3^2+\cdots+n^2=\dfrac{n(n+1)(2 n+1)}{6}$

Hence $I=\displaystyle\lim_{n\to \infty}\dfrac{1}{n}\bigg(\dfrac{1}{n^2}\times \dfrac{(n-1)n(2 n-1)}{6}+\dfrac{1}{n}\times \dfrac{(n-1)n}{2}\bigg)$

$=\displaystyle\lim_{n\to \infty}\bigg(\dfrac{1}{n^2}\times \dfrac{(n-1)(2 n-1)}{6}+\dfrac{1}{n}\times \dfrac{(n-1)}{2}\bigg)$

$=\displaystyle\lim_{n\to \infty}\bigg(\dfrac{1}{6} \bigg(1-\dfrac{1}{n}\bigg)\bigg(2-\dfrac{1}{n}\bigg)+\dfrac{1}{2}\bigg(1-\dfrac{1}{n}\bigg)\bigg)$

$=\dfrac{1}{6}(1-0)(2-0)+\dfrac{1}{2}(1-0)$

$=\dfrac{1}{3}+\dfrac{1}{2}$

$=\dfrac{5}{6}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \lim_{n \rightarrow \infty} \left[\displaystyle \frac{\sqrt{n^{2}-1^{2}}}{n^{2}}+\frac{\sqrt{n^{2}-2^{2}}}{n^{2}}+\frac{\sqrt{n^{2}-3^{2}}}{n^{2}}+\ldots.n terms\right]=$
• A. $\displaystyle \frac{\pi}{2}$
• B. $\displaystyle \frac{\pi}{3}$
• C. $\displaystyle \frac{2\pi}{4}$
• D. $\displaystyle \frac{\pi}{4}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
If $I=\displaystyle \int_{0}^{\pi/2}sinx.log(sin x)dx = log\left(\dfrac{K}{e}\right).$ Then find the value of $K.$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
$\int { \dfrac { { e }^{ x }dx }{ \sqrt { { e }^{ 2x }-1 } } }$

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Q4 Subjective Medium
Solve:
$\displaystyle \int_{0}^{1} \cfrac{2 x+3}{5 x^{2}+1} d x$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
If $y=\displaystyle\int \dfrac {dx}{(1+x^{2})^{\frac {1}{2}}}$ and $y=0$ when $x=0$, then value of $y$ when$x=1$, is:
• A. $\ln(2)$
• B. $\ln(\sqrt{2})$
• C. $\dfrac {1}{\sqrt {2}}$
• D. $\ln(1+\sqrt{2})$