Question 946b3

Jan 28, 2017

$\int \left({x}^{\frac{1}{2}} + 3 x + 5\right) \mathrm{dx} = \frac{2}{3} {x}^{\frac{3}{2}} + \frac{3}{2} {x}^{2} + 5 x + \text{constant}$

Explanation:

Use the power rule.

For $n \ne - 1$,

$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left\{n + 1\right\} + \text{constant}$.

Therefore,

$\int \left({x}^{\frac{1}{2}} + 3 x + 5\right) \mathrm{dx} = \int {x}^{\frac{1}{2}} \mathrm{dx} + 3 \int x \mathrm{dx} + 5 \int {x}^{0} \mathrm{dx}$

$= \frac{2}{3} {x}^{\frac{3}{2}} + \frac{3}{2} {x}^{2} + 5 x + \text{constant}$

Jan 28, 2017

$\frac{2}{3} {x}^{\frac{3}{2}} + \frac{3}{2} {x}^{2} + 5 x + c$

Explanation:

This is a Calculus question.

Integrate each term using the $\textcolor{b l u e}{\text{power rule for integration}}$

color(red)(bar(ul(|color(white)(2/2)color(black)(intax^ndx=a/(n+1)x^(n+1) ; n≠-1)color(white)(2/2)|)))#

$\textcolor{b l u e}{\text{Note that " 5=5x^0" since }} {x}^{0} = 1$

$\Rightarrow \int \left({x}^{\frac{1}{2}} + 3 x + 5\right) \mathrm{dx}$

$= \frac{1}{\frac{3}{2}} {x}^{\left(\frac{1}{2} + 1\right)} + \frac{3}{2} {x}^{\left(1 + 1\right)} + 5 {x}^{\left(0 + 1\right)} + c$

$= \frac{2}{3} {x}^{\frac{3}{2}} + \frac{3}{2} {x}^{2} + 5 x + c$

where c is the constant of integration.