## ta123 Group Title need help on the attachment @Calculus1 2 years ago 2 years ago

1. ta123 Group Title

2. across Group Title

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

3. ta123 Group Title

how come but side are the same

4. across Group Title

What do you mean?

5. ta123 Group Title

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

6. across Group Title

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

7. ta123 Group Title

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

8. across Group Title

They're subtracting.

9. across Group Title

$\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 Group Title

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

11. ta123 Group Title

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