Integral question :O?

Question

Integrate: x((x)^1/3 + (x)^1/4) in the interval [0, 1]. I can solve it without the x on the outside of the equation. However, that pesky x outside is what is throwing me off :O Thanks!

Answer 1

You just need to distribute the x through the rest of the equation:

x (x^1/3 + x^1/4) = x^4/3 + x^5/4

Now, I'm sure you can integrate that yourself, but you should get:

(3/7)x^7/3 + (4/9)x^9/4

Evaluating from [0,1], you just have:

(3/7)*1^7/3 + (4/9)*1^9/4 - 0 = 3/7 + 4/9 = 55/63

Answer 2

You have stated that you can solve it without the x outside. x((x)^1/3+(x)^1/4).
Multiply each term by x, x times x^1/3 = x^(1+1/3)=x^(4/3) and x times x^(1/4)= x^(5/4)
you have two integrals (x)^(4/3)+(x)^(5/4)
integrate each between limits 0 to 1. If my computations are correct, i get 3/7+4/9 which you can add and simplify.

Answer 3

Just multiply through to get:

x^(4/3) + x^(5/4)

integrate:

(3/7)x^(7/3) + (4/9)x^(9/4)

Which is 0 at x=0 and so the answer is simply what you get when you plug in x=1:

3/7+4/9 = 55/63

Answer 4

integ x [ x^1/3 + x^1/4 ]
= integ [ x^4/3 + x^5/4 ]
= [ 3 / 7 x^7/3 + 4 / 9 x^9/4 ] limit [ 0, 1 ]
= [ 3 / 7 + 4 / 9 ]
= ( 27 + 28 ) / 63
= 55 / 63

Answer 5

> I can solve it without the x on the outside of the equation.

Then move it to the inside! Use the distributive property to muliply the x by both of the terms in the parentheses. Then use the law of exponents to simplify x·x^(1/3); and to simplify x·x^(1/4).

Answer 6

integral of[x(x^1/3+()x^1/4]dx
=integral of [x^4/3 +x^5/4]dx
=3/7x^^7/3+4/9x^9/4. with limits 0to1
=3/7+4/9=55/63. answer