# How do I evaluate int_1^4sqrt(t) \(5 + 2 t) dt?

Jan 28, 2015

I would expand the argument as:

${\int}_{1}^{4} \sqrt{t} \cdot 5 + \sqrt{t} \cdot 2 t \mathrm{dt} =$

I can now separate the integrals, as:

${\int}_{1}^{4} \sqrt{t} \cdot 5 \mathrm{dt} + {\int}_{1}^{4} \sqrt{t} \cdot 2 t \mathrm{dt} =$

Remembering that $\sqrt{t} = {t}^{\frac{1}{2}}$ and the rules to manipulate exponentes, to get:

$= {\int}_{1}^{4} {t}^{\frac{1}{2}} \cdot 5 \mathrm{dt} + {\int}_{1}^{4} {t}^{\frac{1}{2} + 1} \cdot 2 \mathrm{dt} =$
$= {\int}_{1}^{4} {t}^{\frac{1}{2}} \cdot 5 \mathrm{dt} + {\int}_{1}^{4} {t}^{\frac{3}{2}} \cdot 2 \mathrm{dt} =$

Taking the constant out of the integrals and remembering that the integral of ${x}^{n}$ is ${x}^{n + 1} / \left(n + 1\right)$;

$5 {t}^{\frac{1}{2} + 1} / \left(\frac{1}{2} + 1\right) + 2 {t}^{\frac{3}{2} + 1} / \left(\frac{3}{2} + 1\right) {|}_{1}^{4}$
$\frac{10}{3} {t}^{\frac{3}{2}} + \frac{4}{5} {t}^{\frac{5}{2}} {|}_{1}^{4} = 48 , 13$