Question

Evaluate the following definite integrals as limit of sums.
${\int }_1^4(x^2-x)dx$.

• A. $\frac{81}{2}$
• B. $\frac{63}{3}$
• C. $\frac{35}{3}$
• D. $\frac{27}{2}$

Answer 1

The correct option is D $\frac{27}{2}$

Here, the function $f\left(x\right)=x^2-x$ is continuous in $[1,4]$. Since dividing the n length of $[1,4]$ in the same lengthg, the length of each subset is h.

i.e., $h=\frac{b-a}{n}=\frac{4-1}{n}=\frac{3}{n}$, here $a=1$ & $b=4$

Now, $f(a+ih)=f(1+ih)=(1+ih{)}^2-(1+ih)$

$f(a+ih)=1+2ih+i^2{h}^2-1-ih$

So, $f(a+ih)=i^2{h}^2+ih$

By definition, ${\int }_1^4(x^2-x)dx=\underset{n\to \infty }{lim}h\sum _{i=1}^{n}f(a+ih)$

So, ${\int }_1^4(x^2-x)dx=\underset{n\to \infty }{lim}\left(\frac{3}{n}\right)\sum _{i=1}^n(i^2{h}^2+ih)$

$=\underset{n\to \infty }{lim}\left(\frac{3}{n}\right)\left[\sum _{i=1}^n{i}^2{h}^2+\sum _{i=1}^{n}ih\right]$

$=\underset{n\to \infty }{lim}\left(\frac{3}{n}\right){\left(\frac{3}{n}\right)}^2\sum _{i=1}^n{i}^2+\underset{n\to \infty }{lim}\left(\frac{3}{n}\right)\left(\frac{3}{n}\right)\sum _{i=1}^{n}i$

$=\underset{n\to \infty }{lim}{\left(\frac{3}{n}\right)}^3\frac{\left(n\right)(n+1)(2n+1)}{6}+\underset{n\to \infty }{lim}{\left(\frac{3}{n}\right)}^2\frac{\left(n\right)(n+1)}{2}$

$=\underset{n\to \infty }{lim}[\frac{(3{)}^3}{6}\left(\frac{n_1}{n}\right)\left(\frac{n+1}{n}\right)\left(\frac{2n+1}{n}\right)+\frac{(3{)}^2}{2}\left(\frac{n}{n}\right)\left(\frac{n+1}{n}\right)]$

$=\underset{n\to \infty }{lim}\left[\left(\frac{27}{6}\right)\right(1)(1+\frac{1}{n})(2+\frac{1}{n})+\frac{9}{2}(1\left)(1+\frac{1}{n})\right]$

${\int }_1^4(x^2-x)dx=\frac{27}{6}\left(1\right)(1+0)(2+0)+\frac{9}{2}\left(1\right)(1+0)=\frac{27}{2}$.