Question

Evaluate the definite integral.
The integral from 0 to a of x*sqrt(a^2-x^2)dx

Answer 1

\[\int\limits_{0}^{a}x \sqrt{a^2-x^2}dx\]

Answer 2

Have you considered a substitution?

Answer 3

Yeah. I set u=a^2-x^2

Answer 4

Yeah, that can work. Or trig.

Answer 5

Okay. Next? du = ??

Answer 6

I don't know trig subsitution.

Answer 7

du=2a-2xdu

Answer 8

Or would a^2 just be left out because it is a constant?

Answer 9

Close.

a is a constant. That should go away.
Why is there a "du" on the right side?

Give it another go.

Answer 10

du=-2xdx
then you would divde by -2 and get
-1/2du=xdx

Answer 11

Okay. Next...

Answer 12

=\[=\int\limits_{0}^{a}\frac{ -1 }{ 2 }(\sqrt{u})du\]

Answer 13

No good. The limits are still x-limits. You have two options:

1) Change the limits to limits of "u", or
2) Solve the indefinite integral and change the variable back to x before evaluating the definite integral.

Answer 14

I solved the indefinite integral and changed it back to x. I got
\[\frac{ 1 }{ 2 }(\frac{ 2 }{ 3 }(a^2-x^2)^{\frac{ 3 }{ 2 }}+C)\]

Answer 15

Then wouldn't you just evaluate it at a and 0? then subtract where you evaluated it at 0 from where you evaluated it at a?

Answer 16

That seems mostly okay. Simplify the expression and figure out where the negative sign went.

Answer 17

Oh, I just forgot to type it. It's still there in my work.

Answer 18

I got \[\frac{ -1 }{ 3 }a^\frac{ 4 }{ 3 }\] as my final answer.

Answer 19

It should not be negative.
How did the 3 int he exponent get into the denominator of the exponent?

Try again. The first term is zero. You should get only the second term.

Answer 20

Reality check. Look at the original argument. For positive 'a', can you find anything negative about that expression? It better not be a negative result.

Answer 21

so would it be \[\frac{ 1 }{ 3 }a^3\]

Answer 22

Why is it NOT \(\dfrac{1}{3}|a|^{3}\)?