Question

Question #26a54

Answer 1

`12/5root3(x^5)+10/root5(x^4)+C`

Explanation:

Rewrite using fractional and negative exponents:

`=int(4x^(2/3)-8x^(-9/5))dx`

Use the rule `intx^ndx=x^(n+1)/(n+1)+C`, where `n!=-1`:

`=4(x^(5/3)/(5/3))-8(x^(-4/5)/(-4/5))+C`

`=12/5x^(5/3)+10x^(-4/5)+C`

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

`=12/5root3(x^5)+10/root5(x^4)+C`

Answer 2

The integral equals `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:
`int4root(3)(x)^2dx-int8/(root(5)(x)^9)dx`

Now take out the constant terms (the things without an `x`):
`4introot(3)(x)^2dx-8int1/(root(5)(x)^9)dx`

Think back to algebra and your exponent rules. How do you express `root(a)(x)^b` using exponents? The rule is:
`root(a)(x)^b=x^(b/a)`

For our problem, this means
`root(3)(x)^2=x^(2/3)`
`1/(root(5)(x)^9)=1/x^(9/5)=x^(-9/5)`

(Also recall that `1/x^a = x^(-a)`).

Now we have:
`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 `4x^3`, for example, is
`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,
`intx^adx=1/(a+1)(x^(a+1))+C`

That means
`4intx^(2/3)dx=4(1/(2/3+1))(x^(2/3+1))=4*3/5*x^(5/3)=12/5x^(5/3)`
`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:
`4intx^(2/3)dx-8intx^(-9/5)dx`
`=12/5x^(5/3)-(-10x^(-5/4))`

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