Question

What is the integral of x squrt (x^2 + a^2) dx integrated from "zero" to "a" where a>0. The answer is [(1/3) times (2squrt2-1) times a^3]. I need help with the process & the steps to reach this answer. Thank you in advance! :)

Answer 1

let x = (a tan(tetha))

Answer 2

Thank you RadEn, but what does tetha mean?

Answer 3

I believe a u-substitution will suffice here, no need for the trig-substitution...:
\(u = x^2 + a^2\)
\(du = 2x \; dx \implies \frac{1}{2} du = x \; dx \)
There is an additional \(x\) multiplied on the outside, so it should work out here in this case.

Answer 4

That's what I did, but I ended up getting (2/6) times (x^2 +a^2)^(3/2) ...ill i have to do next is plug in the "zero ", get an answer, and then plug in the "a" and get an anwer then subtract the two to get an answer...but for some reason, i ended up with (1/3) times a^3..what do you think is the mistake? Thank you so Much AccessDenied!! Your answer was so helpful :))

Answer 5

Hmm...
\[\frac{1}{3}(x^2 + a^2)^{3/2}\]
What do you get when you put x=a?
\(\frac{1}{3}(a^2 + a^2)^{3/2}\)
and x=0:
\(\frac{1}{3}(0^2 + a^2)^{3/2}\)

Answer 6

i got it....
but very long time to solve it... :(

Answer 7

\[ \frac{1}{3}(2a^\cancel{2})^{3/\cancel{2}} \\
= \frac{1}{3} (2)^{3/2} a^3 \\
= \frac{1}{3} 2 \sqrt{2} \; a^3 \] \[
\frac{1}{3}(a^\cancel{2})^{3\cancel{/2}} \\
\frac{1}{3}a^3
\]

Answer 8

When we subtract the x=a from the x=0, we get:
\[ \frac{1}{3} 2\sqrt{2} \; a^3 - \frac{1}{3} a^3 \\
= \frac{1}{3} a^3 \left( 2 \sqrt{2} - 1 \right) \]
Which could be rearranged to fit the answer given.

Answer 9

* subtract the value of x=0 from x=a, that would be more appropriate to say... :P