U substitution

  • #1

Homework Statement


Im looking over the notes in my lecture and the prof wrote,
[tex] \int_{0}^{2} \pi(4x^2-x^4)dx=\frac{64\pi}{15} [/tex]
Im wondering whats the indefinite integral of this equation.

Homework Equations


using u substitution


The Attempt at a Solution


[tex] \int \pi(4x^2-x^4)dx= \pi \int x^2(4-x^2)dx \\
u = 4 - x^2 \ \ \ \ \ \ \ \ \ \
-\frac {1}{2} du =xdx \\
[/tex]

Im confuse since i have an x^2 but my du=x.

I attempted to also use from u to get,
[tex]
x=\sqrt{4-u} \\
\pi \int \frac{1}{2}u\sqrt{4-u}dx
[/tex]
but it seems this made the formula harder to integrate...or am i just giving up too quickly
 

Answers and Replies

  • #2
Samy_A
Science Advisor
Homework Helper
1,241
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Homework Statement


Im looking over the notes in my lecture and the prof wrote,
[tex] \int_{0}^{2} \pi(4x^2-x^4)dx=\frac{64\pi}{15} [/tex]
Im wondering whats the indefinite integral of this equation.

Homework Equations


using u substitution


The Attempt at a Solution


[tex] \int \pi(4x^2-x^4)dx= \pi \int x^2(4-x^2)dx \\
u = 4 - x^2 \ \ \ \ \ \ \ \ \ \
-\frac {1}{2} du =xdx \\
[/tex]

Im confuse since i have an x^2 but my du=x.

I attempted to also use from u to get,
[tex]
x=\sqrt{4-u} \\
\pi \int \frac{1}{2}u\sqrt{4-u}dx
[/tex]
but it seems this made the formula harder to integrate...or am i just giving up too quickly
You are making this far too complicated.

What is ##\int x^n dx##, (for ##n \geq 1##)?
 
  • #3
That makes sense...

I was working on a bunch of u sub equations earlier. I guess I was on a u sub mode

Thanks a lot!
 

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