Evaluate the iterated integral below.Note that it is necessary to switch the order of integration.

Question

Quizplus · Accepted Answer

We can switch the order of integration by writing the bounds of integration for x and y. We have:

∫[0, 1] ∫[0, x^2] dy dx ∫[0, y] f(x, y) dz

Switching the order of integration gives:

∫[0, 1] ∫[sqrt(y), 1] dx dy ∫[0, x^2] f(x, y) dz

We can then evaluate the iterated integral as:

∫[0, 1] ∫[sqrt(y), 1] ∫[0, x^2] e^(x^2 + y^2) dz dx dy

Using the properties of exponential functions, we can evaluate the innermost integral to get:

∫[0, x^2] e^(x^2 + y^2) dz = e^(x^2 + y^2) ∣[0, x^2] = e^(x^2)(e^y - 1)

Substituting back into the iterated integral, we get:

∫[0, 1] ∫[sqrt(y), 1] e^(x^2)(e^y - 1) dx dy

We can now evaluate the integrals one by one, starting with the integral with respect to x:

∫[sqrt(y), 1] e^(x^2)(e^y - 1) dx = (e^y - 1) ∫[sqrt(y), 1] e^(x^2) dx

This integral does not have an elementary antiderivative, so we can write it in terms of the error function:

(e^y - 1) ∫[sqrt(y), 1] e^(x^2) dx = (e^y - 1) √π/2 erf(sqrt(2)y/2)

Substituting back into the iterated integral, we get:

∫[0, 1] (e^y - 1) √π/2 erf(sqrt(2)y/2) dy

This integral can be evaluated using integration by parts, where u = erf(sqrt(2)y/2) and dv = (e^y - 1) dy:

∫(e^y - 1) √π/2 erf(sqrt(2)y/2) dy = -e^y √π/2 erf(sqrt(2)y/2) + ∫e^y √2/π e^-(sqrt(2)y/2)^2 dy - y √π/2 erf(sqrt(2)y/2)

The first term evaluates to -√π/2, and the second term evaluates to (√2/π)(e^1 - 1). The third term evaluates to 0 at both bounds. Therefore, the value of the iterated integral is:

-√π/2 + (√2/π)(e^1 - 1) = 11ebbd4d_3f0d_c307_8a0a_cf21593a1a2f_TB8202_11

Evaluate the Iterated Integral Below