Question #26a54

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Question #26a54

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Answer 1 · 2017-08-29T02:16:49

$\frac{12}{5} \sqrt[3]{x^{5}} + \frac{10}{\sqrt[5]{x^{4}}} + C$ $12/5root3(x^5)+10/root5(x^4)+C$

Explanation:

Rewrite using fractional and negative exponents:

$= \int (4 x^{\frac{2}{3}} - 8 x^{- \frac{9}{5}}) d x$ $=int(4x^(2/3)-8x^(-9/5))dx$

Use the rule $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $intx^ndx=x^(n+1)/(n+1)+C$ , where $n \neq - 1$ $n!=-1$ :

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

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

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

$= \frac{12}{5} \sqrt[3]{x^{5}} + \frac{10}{\sqrt[5]{x^{4}}} + C$ $=12/5root3(x^5)+10/root5(x^4)+C$

Answer 2 · 2017-08-29T02:24:58

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

Explanation:

Use the sum rule to break the big integral into littler ones:
$\int 4 {\sqrt[3]{x}}^{2} d x - \int \frac{8}{{\sqrt[5]{x}}^{9}} d x$ $int4root(3)(x)^2dx-int8/(root(5)(x)^9)dx$

Now take out the constant terms (the things without an $x$ $x$ ):
$4 \int {\sqrt[3]{x}}^{2} d x - 8 \int \frac{1}{{\sqrt[5]{x}}^{9}} d x$ $4introot(3)(x)^2dx-8int1/(root(5)(x)^9)dx$

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

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

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

Now we have:
$4 \int x^{\frac{2}{3}} d x - 8 \int x^{- \frac{9}{5}} d x$ $4intx^(2/3)dx-8intx^(-9/5)dx$

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 $4 x^{3}$ $4x^3$ , for example, is
$3 \cdot 4 x^{3 - 1} = 12 x^{2}$ $3*4x^(3-1)=12x^2$

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,
$\int x^{a} d x = \frac{1}{a + 1} (x^{a + 1}) + C$ $intx^adx=1/(a+1)(x^(a+1))+C$

That means
$4 \int x^{\frac{2}{3}} d x = 4 (\frac{1}{\frac{2}{3} + 1}) (x^{\frac{2}{3} + 1}) = 4 \cdot \frac{3}{5} \cdot x^{\frac{5}{3}} = \frac{12}{5} x^{\frac{5}{3}}$ $4intx^(2/3)dx=4(1/(2/3+1))(x^(2/3+1))=4*3/5*x^(5/3)=12/5x^(5/3)$
$8 \int x^{- \frac{9}{5}} d x = 8 (\frac{1}{- \frac{9}{5} + 1}) (x^{- \frac{9}{5} + 1}) = 8 \cdot - \frac{5}{4} \cdot x^{- \frac{5}{4}} = - 10 x^{- \frac{5}{4}}$ $8intx^(-9/5)dx=8(1/(-9/5+1))(x^(-9/5+1))=8*-5/4*x^(-5/4)=-10x^(-5/4)$

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

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

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Question #26a54

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