Mathematics

# Evaluate the following definite integrals as limit of sums.$\displaystyle\int^5_0(x+1)dx$.

$\dfrac{35}{2}$

##### SOLUTION
Here the function $f(x)=x+1$ is continuous in $[0, 5]$. Since dividing the n length of $[0, 5]$ in the same length of each subset is h.
i.e., $h=\dfrac{b-a}{n}=\dfrac{5-0}{n}=\dfrac{5}{n}$, here $a=0$ & $b=5$
$f(a+ih)=f(ih+0)=f(ih)=ih+1$
By definition $\displaystyle\int^5_0(x+1)dx=\displaystyle\lim_{n\rightarrow \infty}h\displaystyle\sum^n_{i=1}f(a+ih)$
So, $\displaystyle\int^5_0(x+1)dx=\displaystyle\lim_{n\rightarrow \infty}\left(\dfrac{5}{n}\right)\displaystyle\sum^n_{i=1}(ih+1)$
$=\displaystyle\lim_{n\rightarrow \infty}\left(\dfrac{5}{n}\right)\left[h\displaystyle\sum^n_{i=1}i+\displaystyle\sum^n_{i=1}1\right]$
$=\displaystyle\lim_{n\rightarrow \infty}\left(\dfrac{5}{n}\right)\left[\left(\dfrac{5}{n}\right)\left(\dfrac{n(n+1)}{2}\right)+(n)\right]$
$=\displaystyle\lim_{n\rightarrow \infty}\left[\dfrac{25}{2}\left(\dfrac{n}{n}\right)\left(\dfrac{n+1}{n}\right)+5\left(\dfrac{n}{n}\right)\right]$
$=\displaystyle\lim_{n\rightarrow \infty}\left[\dfrac{25}{2}(1)\left(1+\dfrac{1}{n}\right)+5(1)\right.$
$=\displaystyle\lim_{n\rightarrow \infty}\dfrac{25}{2}\left(1+\dfrac{1}{h}\right)+\displaystyle\lim_{n\rightarrow \infty}5$
$=\dfrac{25}{2}(1+0)+5$
$=\dfrac{25}{2}+5$
$\displaystyle\int^5_0(x+1)dx=\dfrac{35}{2}$.

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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