integrals

[jeca86]

Evaluate the definite integral.

the letter "a" is at the top of the integral and 0 is at the bottom. the rest of the equation is x*square root of (a^2 -x^2)dx

so i have tried to do this but i get really confused. i know i have to solve by substitution first. i think this is how you start:

u=a^2-x^2, so, x=square root(-u+a^2)
du=(a^2-2x)dx, so dx=du/(a^2-2x)
when x=a, u=0
when x=0, u=a^2

im not sure if this is right. can someone help me get through the rest?

 

[Unco]

a^2 is just a constant so du = -2x dx.

 

[soroban]

Hello, jeca86!

Evaluate: \(\displaystyle \L\,\int^{\;\;\;a}_0x\cdot\sqrt{a^2\,-\,x^2}\,dx\)

Your work is correct . . . I prefer to do it this way:

LetL \(\displaystyle u\:=\:a^2\,-\,x^2\;\;\Rightarrow\;\;du\:=\:-2x\,dx\;\;\Rightarrow\;\;dx\:=\:-\frac{du}{2x}\)

When \(\displaystyle x\,=\,a,\:u\,=\0\)
When \(\displaystyle x\,=\,a,\:u\,=\,a^2\)

Substitute: \(\displaystyle \L\:\int^{\;\;\;0}_{a^2}x\cdot u^{1/2}\cdot\left(-\frac{du}{2x}\right)\;=\;-\frac{1}{2}\int^{\;\;\;0}_{a^2}u^{\frac{1}{2}}\,du\;=\;-\frac{1}{2}\cdot\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\,\bigg]^0_{a^2}\)

\(\displaystyle \L\;\;\;= \;-\frac{1}{3}u^{\frac{3}{2}}\,\bigg]^0_{a^2}\;=\;\left[-\frac{1}{3}\cdot0^{\frac{3}{2}}\right]\,-\,\left[-\frac{1}{3}\cdot(a^2)^{\frac{2}{3}}\right] \;= \;\frac{1}{3}a^3\)

 

[jeca86]

thank you