Question

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

• A. $\frac{15}{2}$
• B. $\frac{25}{2}$
• C. $\frac{35}{2}$
• D. $\frac{45}{2}$

Answer 1

The correct option is C $\frac{35}{2}$

Here the function $f\left(x\right)=x+1$ is continuous in $[0,5]$. Since dividing the n length of $[0,5]$ in the same length of each subset is h.

i.e., $h=\frac{b-a}{n}=\frac{5-0}{n}=\frac{5}{n}$, here $a=0$ & $b=5$

$f(a+ih)=f(ih+0)=f\left(ih\right)=ih+1$

By definition ${\int }_0^5(x+1)dx=\underset{n\to \infty }{lim}h\sum _{i=1}^{n}f(a+ih)$

So, ${\int }_0^5(x+1)dx=\underset{n\to \infty }{lim}\left(\frac{5}{n}\right)\sum _{i=1}^n(ih+1)$

$=\underset{n\to \infty }{lim}\left(\frac{5}{n}\right)\left[h\sum _{i=1}^{n}i+\sum _{i=1}^{n}1\right]$

$=\underset{n\to \infty }{lim}\left(\frac{5}{n}\right)[\left(\frac{5}{n}\right)\left(\frac{n(n+1)}{2}\right)+(n\left)\right]$

$=\underset{n\to \infty }{lim}[\frac{25}{2}\left(\frac{n}{n}\right)\left(\frac{n+1}{n}\right)+5\left(\frac{n}{n}\right)]$

$=\underset{n\to \infty }{lim}\left[\frac{25}{2}\right(1)(1+\frac{1}{n})+5(1)$

$=\underset{n\to \infty }{lim}\frac{25}{2}(1+\frac{1}{h})+\underset{n\to \infty }{lim}5$

$=\frac{25}{2}(1+0)+5$

$=\frac{25}{2}+5$

${\int }_0^5(x+1)dx=\frac{35}{2}$.