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# How do I reverse this integral and evaluate it?

I could really use some help for the following integral:

∫ (0 to 9) ∫ (sqrt(y) to 3) sin(x^3) dxdy

I understand I need to sketch the region of the area first to find its new bounds. When I did, I got ∫ (0 to 3) ∫ (x^2 to 9) sin(x^3) dydx. Can you help my verify if that's right or wrong? Also, how do I solve this integral after it is reversed? I could use some help. Thank you in advance.

### 1 Answer

- kbLv 78 years agoFavorite Answer
The bounds of integration are x = sqrt(y) to x =3 with y in [0, 9].

Since x = sqrt(y) <==> y = x^2, we can rewrite this more nicely as

y = 0 to y = x^2 with x in [0, 3]. (Sketching the region is useful to see this.)

So, we obtain

∫(x = 0 to 3) ∫(y = 0 to x^2) sin(x^3) dy dx

= ∫(x = 0 to 3) y sin(x^3) {for y = 0 to x^2} dx

= ∫(x = 0 to 3) x^2 sin(x^3) dx

= (-1/3) cos(x^3) {for x = 0 to 3}

= (1/3)(1 - cos 27).

I hope this helps!