Evaluate the definite integral

[frosty8688]

1. $\int^{a}_{0} x\sqrt{x^{2}+a^{2}}$ a > 0



2. u = x$^{2}$ + a$^{2}$, du = 2x



3. $\frac{1}{2}\int^{a}_{0}\frac{2u^{3/2}}{3} = \frac{1}{3}\int ^{a}_{0}(x^{2}+a^{2})^{3/2}$ How do I solve this? Any hints?

 

[Mark44]

1. $\int^{a}_{0} x\sqrt{x^{2}+a^{2}}$ a > 0



2. u = x$^{2}$ + a$^{2}$, du = 2x



3. $\frac{1}{2}\int^{a}_{0}\frac{2u^{3/2}}{3} = \frac{1}{3}\int ^{a}_{0}(x^{2}+a^{2})^{3/2}$ How do I solve this? Any hints?

You have omitted dx. You might think this to be an unimportant appendage that can be ignored, but as you learn additional integration techniques, it becomes very important.

In your ordinary substitution, u = x² + a², and du = 2xdx.

Making the substitution, you get
$$ \frac{1}{2}\int_0^a u^{1/2}du = \frac{2}{2 \cdot 3}\left. u^{3/2} \right|_0^a$$
$$ =\frac{1}{3} \left. (\sqrt{x^2 + a^2})^3 \right|_0^a$$

Evaluate the quantity at x = a, and then at x = 0, and subtract the two values.

 

[frosty8688]

So we let a = 1 and 1+1 = 2 and it would be sqrt 2^3 which would be 2 sqrt 2 - (0+1)^3 = 1 all to a^3 * 1/3.

 

[Mark44]

So we let a = 1 and 1+1 = 2 and it would be sqrt 2^3 which would be 2 sqrt 2 - (0+1)^3 = 1 all to a^3 * 1/3.

No, you don't assign a value to a.

Instead of writing stuff like "which would be 2 sqrt 2 - (0+1)^3 = 1 all to a^3 * 1/3" that makes little sense, try writing some actual mathematics, with = between equal expressions.

 

[frosty8688]

So $\left|x\right| = a$ and -a.

 

[frosty8688]

Is there an equation I can use to compute the values of x?

 

[frosty8688]

I got it when a is evaluated x = +/- 1 and since x=a +/- 1^2 + +/- 1^2 = 2 and when the lower limit is evaluated x = 0, but a > 0 so that makes the expression equal to 1.

 

[SammyS]

I got it when a is evaluated x = +/- 1 and since x=a +/- 1^2 + +/- 1^2 = 2 and when the lower limit is evaluated x = 0, but a > 0 so that makes the expression equal to 1.

"a" is simply "a", an arbitrary positive real number. Your answer should have "a" in it.

 

[frosty8688]

I have $\frac{1}{3}(2\sqrt{2}-1)a^{3}$

 

[SammyS]

I have $\frac{1}{3}(2\sqrt{2}-1)a^{3}$

Looks good.