Calculus Integral.

[Assassin315]

I've already done what I presume to be the main part of this problem, and I'm left with proving the following:

This probably isn't as hard as I'm making it, but maybe someone will quickly correct me where I've gone wrong.

I set u = a^(2/3) - x^(2/3), meaning du = (-2/3)x^(-1/3) dx, or du = -2/(3*(x^(1/3))) dx. Thus, dx = (3*x^(1/3))/2 du.

This leaves me with the integral of u^3 du over the specified boundary. The integral of u^3 is u^4/4 -- and so (a^(2/3) - x^(2/3)^4)/4) after subbing u back in -- and I multiplied that by my du. In the end, I got a simplified version. However, if I try to integrate it from 0 to a, my bounds are either equal to 0 or... don't exist. What'd I mess up on?

 

[Deleted member 4993]

I've already done what I presume to be the main part of this problem, and I'm left with proving the following:

This probably isn't as hard as I'm making it, but maybe someone will quickly correct me where I've gone wrong.

I set u = a^(2/3) - x^(2/3), meaning du = (-2/3)x^(-1/3) dx, or du = -2/(3*(x^(1/3))) dx. Thus, dx = (3*x^(1/3))/2 du.

This leaves me with the integral of u^3 du over the specified boundary. The integral of u^3 is u^4/4 -- and so (a^(2/3) - x^(2/3)^4)/4) after subbing u back in -- and I multiplied that by my du. In the end, I got a simplified version. However, if I try to integrate it from 0 to a, my bounds are either equal to 0 or... don't exist. What'd I mess up on?

What did you do with that part?

substitute

x^(2/3) = a^(2/3) * sin[sup:k367sn2m]2[/sup:k367sn2m](?)

x^(1/3) = a^(1/3) * sin(?)

1/3 * x^(-2/3) dx = a^(1/3) * cos(?) d?

dx = 3 * x^(2/3) * a^(1/3) * cos(?) d?

dx = 3 * [ a^(2/3) * sin[sup:k367sn2m]2[/sup:k367sn2m](?) ] * a^(1/3) * cos(?) d?

dx = 3 * a * sin[sup:k367sn2m]2[/sup:k367sn2m](?) * cos(?) d?

Then

[a^(2/3) - x^(2/3)][sup:k367sn2m]3[/sup:k367sn2m]dx

= [ a^(2/3) - a^(2/3) * sin[sup:k367sn2m]2[/sup:k367sn2m](?)][sup:k367sn2m]3[/sup:k367sn2m] * 3 * a * sin[sup:k367sn2m]2[/sup:k367sn2m](?) * cos(?) d?

= 3 * a[sup:k367sn2m]3[/sup:k367sn2m]cos[sup:k367sn2m]7[/sup:k367sn2m](?) * sin[sup:k367sn2m]2[/sup:k367sn2m](?) d?

Integrate - sub back and evaluate limits.....

 

[soroban]

Hello, Assassin315!

\(\displaystyle \text{Prove: }\:\int^a_0\left(a^{\frac{2}{3}} - x^{\frac{2}{3}}\right)^3dx \:=\:\frac{16}{105}a^3\)

\(\displaystyle \text{We have: }\;\int^a_0\left(a^2 - 3a^{\frac{4}{3}}x^{\frac{2}{3}} + 3a^{\frac{2}{3}}x^{\frac{4}{3}} - x^2\right)\,dx \;\;=\;\; a^2x \;-\; \tfrac{9}{5}a^{\frac{4}{3}}x^{\frac{5}{3}} \;+\; \tfrac{9}{7}a^{\frac{2}{3}}x^{\frac{7}{3}} \;-\; \tfrac{1}{3}x^3\,\bigg]^a_0\)


\(\displaystyle \text{Evaluate: }\;\bigg[a^2(a) - \tfrac{9}{5}a^{\frac{4}{3}}\left(a^{\frac{5}{3}}\right) + \tfrac{9}{7}a^{\frac{2}{3}}\left(a^{\frac{7}{3}}\right) - \tfrac{1}{3}a^3\bigg] - \bigg[ 0 - 0 + 0 - 0\bigg]\)

. . . . . . . . \(\displaystyle =\;\; a^3 - \tfrac{9}{5}a^3 + \tfrac{9}{7}a^3 - \tfrac{1}{3}a^3 \;\;=\;\;\left(\frac{105 - 189 + 135 - 35}{105}\right)a^3 \;\;=\;\;\frac{16}{105}a^3\)

 

[Assassin315]

Ooooh, it makes so much more sense to expand it then to do some bizarre u-substitution. I realized too earlier today that you can actually use the binomial theorem to quickly expand it if needed, but I didn't realize how simple it was afterwards.

Thanks!