Evaluate the Integral. Show All Steps

Question

Evaluate the Integral. Show All Steps. If possible attach a video going through the problem.

Simplify by factoring then check by differentiation.

Answer 1

We will use integration by parts. Recall the formula for it:

∫ u dv = uv - ∫ v du

Let u = (1/3)x³ + 2x² + 1 and dv = e^-x/2 dx.

Then du = (x² + 4x)dx and v = -2e^-x/2.

∫ e^-x/2 [(1/3)x³ + 2x² + 1] dx

= -2e^-x/2 [(1/3)x³ + 2x² + 1 ] - ∫ -2e^-x/2 (x² + 4x) dx

= e^-x/2 [(-2/3)x³ - 4x² - 2] + ∫ 2e^-x/2 (x² + 4x) dx

= e^-x/2 [(-2/3)x³ - 4x² - 2] + ∫ e^-x/2 (2x² + 8x) dx

We use integration by parts one more time.

Let u = 2x² + 8x and dv = e^-x/2 dx.

Then du = (4x + 8)dx and v = -2e^-x/2.

∫ e^-x/2 [(1/3)x³ + 2x² + 1] dx

= e^-x/2 [(-2/3)x³ - 4x² - 2] + ∫ e^-x/2 (2x² + 8x) dx

= e^-x/2 [(-2/3)x³ - 4x² - 2] + (-2e^-x/2)(2x² + 8x) - ∫ -2e^-x/2 (4x + 8) dx

= e^-x/2 [(-2/3)x³ - 4x² - 2] + e^-x/2(-4x² - 16x) + ∫ 2e^-x/2 (4x + 8) dx

= e^-x/2 [(-2/3)x³ - 4x² - 2 - 4x² - 16x] + ∫ e^-x/2 (8x + 16) dx

= e^-x/2 [(-2/3)x³ - 8x² - 16x - 2] + ∫ e^-x/2 (8x + 16) dx

We use integration by parts one more time.

Let u = 8x + 16 and dv = e^-x/2 dx.

Then du = 8 dx and v = -2e^-x/2.

∫ e^-x/2 [(1/3)x³ + 2x² + 1] dx

= e^-x/2 [(-2/3)x³ - 8x² - 16x - 2] + ∫ e^-x/2 (8x + 16) dx

= e^-x/2 [(-2/3)x³ - 8x² - 16x - 2] + (-2e^-x/2)(8x + 16) - ∫ (-2e^-x/2)(8) dx

= e^-x/2 [(-2/3)x³ - 8x² - 16x - 2] + (e^-x/2)(-16x - 32) + ∫ (2e^-x/2)(8) dx

= e^-x/2 [(-2/3)x³ - 8x² - 16x - 2 - 16x - 32] + 16 ∫ e^-x/2 dx

= e^-x/2 [(-2/3)x³ - 8x² - 32x - 34] + 16(-2e^-x/2) + C

= e^-x/2 [(-2/3)x³ - 8x² - 32x - 34] -32 e^-x/2 + C

= e^-x/2 [(-2/3)x³ - 8x² - 32x - 34 - 32] + C

= e^-x/2 [(-2/3)x³ - 8x² - 32x - 66] + C

If we factor out a (-2/3), the answer would look a little nicer. In that case, we get:

(-2/3)e^-x/2 [x³ + 12x² + 48x + 99] + C

because

(-2/3) * 12 = - 8

(-2/3) * 48 = -32

(-2/3) * 99 = -66

Detailed explanation of ∫ e^-x/2 dx:

We use u-substitution.

Let u = -x/2.

Then du = (-1/2)dx

du = (-1/2)dx ⇒ -2du = (-2)(-1/2)dx ⇒ -2du = dx

∫ e^-x/2 dx = ∫ e^u (-2du) = -2 ∫ e^u du = -2e^u + C = -2e^-x/2 + C