## ta123 Group Title need help on the attachment @Calculus1 3 years ago 3 years ago

1. ta123

2. across

$\int_{0}^{1}\pi(\sqrt[4]{x})^2dx-\int_{0}^{1}\pi(x^4)^2dx$

3. ta123

how come but side are the same

4. across

What do you mean?

5. ta123

oh, wait my eyes are playing tricks on me I thought I saw a square root in the right integral

6. across

Yeah, integrals will do that to your eyes. :P

7. ta123

I got 7pi/9 is that what you got?

8. across

They're subtracting.

9. across

$\pi\int_{0}^{1}\sqrt{x}dx=\frac{2}{3}\pi\left[x^{\frac{3}{2}}\right]_{0}^{1}=\frac{2}{3}\pi,$$\pi\int_{0}^{1}x^8dx=\pi\left[\frac{1}{9}x^9\right]_{0}^{1}=\frac{1}{5}=\frac{1}{9}\pi,$$\frac{2}{3}\pi-\frac{1}{9}\pi=\frac{5}{9}\pi.$

10. ta123

dang I integrated wrong by different number number to the exponents beside the number 1

11. ta123

oops! I forgot "adding" in between by and different