Question

Compute the following iterated integrals

Answer 1

a)
`(1)/(6)(e^3-1)\ (e^1-1)\\`

b)
`(128)/(35)`

Explanation:

a)

`\int\limits^1_0 {\int\limits^1_0 {x^2(y+1)e^(x^3+y^2+2y)} \, dy } \, dx`

Inside integral is relative to y, so treat x as constant. We can pull x² out of the integral:

`\int\limits^1_0 x^2{\int\limits^1_0 {(y+1)e^(x^3+y^2+2y)} \, dy } \, dx`

Multiply and divide inside integral by 2:

`\int\limits^1_0 (x^2)/(2){\int\limits^1_0 {(2y+2)e^(x^3+y^2+2y)} \, dy } \, dx`

Use identity ∫ eᵘ du = eᵘ.

`\int\limits^1_0 (x^2)/(2)(e^(x^3+y^2+2y)\ |_0^1) \, dx\\\int\limits^1_0 (x^2)/(2)(e^(x^3+3)-e^(x^3)) \, dx\\(1)/(2)(e^3-1) \int\limits^1_0 x^2e^(x^3) \, dx`

Multiply and divide by 3.

`(1)/(6)(e^3-1) \int\limits^1_0 3x^2e^(x^3) \, dx`

Use same identity as before:

`(1)/(6)(e^3-1)\ e^(x^3)|_0^1\\(1)/(6)(e^3-1)\ (e^1-1)\\`

b)

`\int\limits^2_0 {\int\limits^(2x)_(x^2) {x^2y \, dy } \, dx`

Inside integral is relative to y, so treat x as constant. Pull x² out of inner integral:

`\int\limits^2_0 x^2{\int\limits^(2x)_(x^2) {y \, dy } \, dx`

Integrate:

`\int\limits^2_0 x^2((1)/(2) y^2|_(x^2)^(2x)) \, dx\\\int\limits^2_0 (1)/(2)x^2( (2x)^2-(x^2)^2) \, dx\\\int\limits^2_0 (1)/(2)x^2( 4x^2-x^4) \, dx\\(1)/(2)\int\limits^2_0 (4x^4-x^6) \, dx`

Integrate again:

`(1)/(2) ((4)/(5) x^5-(1)/(7) x^7)|_0^2\\(1)/(2) ((4)/(5)\ 2^5-(1)/(7)\ 2^7)\\2^4((4)/(5)-(1)/(7)\ 2^2)\\16((4)/(5)-(4)/(7))\\64((1)/(5)-(1)/(7))\\64((2)/(35))\\(128)/(35)`