$\displaystyle \int_0^1 \int_{2y}^2 \cos(x^2) \, dx \, dy = \int_0^2 \int_0^{x/2} \cos(x^2) \, dy \, dx \\\\ ~~~~~~~~ = \int_0^2 \cos(x^2) y \bigg|_{y=0}^{y=x/2} \, dx \\\\ ~~~~~~~~ = \frac12 \int_0^2 x \cos(x^2) \, dx$

Substitute $u=x^2$ and $du=2x\,dx$ .

$\displaystyle \frac12 \int_0^2 x \cos(x^2) \, dx = \frac14 \int_0^4 \cos(u) \, du = \frac14 \sin(u) \bigg|_{u=0}^{u=4} = \boxed{\frac{\sin(4)}4}$

4 0

1 year ago

Charlotte owns two entertainment websites. Here are some details about those websites for one entire month:

DedPeter [7]

It is d and b I believe

5 0

3 years ago

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Solve for x radical forms

vova2212 [387]

Answer:

in right angled triangle ABC

by using Pythagoras law

h²=p²+b²

x²=7²+10²

x=√149=12.21cm

8 0

3 years ago

Solve the equation 4y-1.9=6.3

Liono4ka [1.6K]

I’ve attached my work...
Hope this helps!

5 0

3 years ago

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