Anonymous

# Help with calculus area problem please :)?

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- 2 months ago
You may find this problem easier if you tilt your head 90 degrees: we're just going to integrate on the y-axis.

Integrate f(y) = y/sqrt(49-y^2) from 0 to 5.

First, let's find the indefinite intregral. We have a function of y^2 in the denominator, and a y on the outside, so let's use substitution.

u = 49 - y^2, du = -2y dy, y dy = -1/2 du

So ∫ y dy/sqrt(49-y^2) = ∫ (-1/2 du)/sqrt(u) = -1/2∫u^-1/2 du

Using the power rule: ∫u^-1/2 du = 2u^1/2 + C

Multiplying by the -1/2 on the outside we have -u^1/2 + C, and by reverse-substituting we get that the indefinite integral is -sqrt(49 - y^2).

Now, evaluate that at 0 and at 5 to get the definite integral.

-sqrt(49 - 5^2) - (-sqrt(49)) = -4.8990 + 7 = 2.101 (to 3 d.p.)

Hope this helped!

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