Mathematics

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


ANSWER

$$\dfrac{35}{2}$$


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
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