# Question #26a54

##### 2 Answers
Aug 29, 2017

$\frac{12}{5} \sqrt[3]{{x}^{5}} + \frac{10}{\sqrt[5]{{x}^{4}}} + C$

#### Explanation:

Rewrite using fractional and negative exponents:

$= \int \left(4 {x}^{\frac{2}{3}} - 8 {x}^{- \frac{9}{5}}\right) \mathrm{dx}$

Use the rule $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$, where $n \ne - 1$:

$= 4 \left({x}^{\frac{5}{3}} / \left(\frac{5}{3}\right)\right) - 8 \left({x}^{- \frac{4}{5}} / \left(- \frac{4}{5}\right)\right) + C$

$= \frac{12}{5} {x}^{\frac{5}{3}} + 10 {x}^{- \frac{4}{5}} + C$

We can write this in the original format if we want:

$= \frac{12}{5} \sqrt[3]{{x}^{5}} + \frac{10}{\sqrt[5]{{x}^{4}}} + C$

Aug 29, 2017

The integral equals $\frac{12}{5} {x}^{\frac{5}{3}} + 10 {x}^{- \frac{5}{4}} + C$; see below for evaluation.

#### Explanation:

Use the sum rule to break the big integral into littler ones:
$\int 4 {\sqrt[3]{x}}^{2} \mathrm{dx} - \int \frac{8}{{\sqrt[5]{x}}^{9}} \mathrm{dx}$

Now take out the constant terms (the things without an $x$):
$4 \int {\sqrt[3]{x}}^{2} \mathrm{dx} - 8 \int \frac{1}{{\sqrt[5]{x}}^{9}} \mathrm{dx}$

Think back to algebra and your exponent rules. How do you express ${\sqrt[a]{x}}^{b}$ using exponents? The rule is:
${\sqrt[a]{x}}^{b} = {x}^{\frac{b}{a}}$

For our problem, this means
${\sqrt[3]{x}}^{2} = {x}^{\frac{2}{3}}$
$\frac{1}{{\sqrt[5]{x}}^{9}} = \frac{1}{x} ^ \left(\frac{9}{5}\right) = {x}^{- \frac{9}{5}}$

(Also recall that $\frac{1}{x} ^ a = {x}^{- a}$).

Now we have:
$4 \int {x}^{\frac{2}{3}} \mathrm{dx} - 8 \int {x}^{- \frac{9}{5}} \mathrm{dx}$

These integrals are easily solved using the reverse power rule. Remember that to differentiate, we brought the power to the front and then reduced the power by one; so the derivative to $4 {x}^{3}$, for example, is
$3 \cdot 4 {x}^{3 - 1} = 12 {x}^{2}$

Since integration is the opposite of differentiation, we do the opposite here: increase the power by one and divide by the new power. In general,
$\int {x}^{a} \mathrm{dx} = \frac{1}{a + 1} \left({x}^{a + 1}\right) + C$

That means
$4 \int {x}^{\frac{2}{3}} \mathrm{dx} = 4 \left(\frac{1}{\frac{2}{3} + 1}\right) \left({x}^{\frac{2}{3} + 1}\right) = 4 \cdot \frac{3}{5} \cdot {x}^{\frac{5}{3}} = \frac{12}{5} {x}^{\frac{5}{3}}$
$8 \int {x}^{- \frac{9}{5}} \mathrm{dx} = 8 \left(\frac{1}{- \frac{9}{5} + 1}\right) \left({x}^{- \frac{9}{5} + 1}\right) = 8 \cdot - \frac{5}{4} \cdot {x}^{- \frac{5}{4}} = - 10 {x}^{- \frac{5}{4}}$

Putting it all together, we have:
$4 \int {x}^{\frac{2}{3}} \mathrm{dx} - 8 \int {x}^{- \frac{9}{5}} \mathrm{dx}$
$= \frac{12}{5} {x}^{\frac{5}{3}} - \left(- 10 {x}^{- \frac{5}{4}}\right)$

For a final answer of $\frac{12}{5} {x}^{\frac{5}{3}} + 10 {x}^{- \frac{5}{4}} + C$. Don't forget the constant of integration $C$!