Evaluate the integral using substitution

Question

1. ∫x²√(2+x)dx
2. 0, 1 ∫(3t-1)⁵⁰dt
3. -π/3, π/3 ∫x⁴sinxdx

Answer 1

∫ x²√(2+x)dx

Let u = 2+x. So, x = u - 2 and dx = du

So we have ∫ (u - 2)²√u du = ∫ [(u² - 4u + 4)√u]du = ∫[u^5/2 - 4u^3/2 + 4u^1/2]du

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∫_{(0 to 1)}(3t - 1)⁵⁰dt

Let u = 3t - 1. Then du = 3dt. So, dt = (1/3)du.

When t = 0, u = -1 and when t = 1, u = 2.

So, we have (1/3)∫_{(-1 to 2)} u⁵⁰du

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sin(-x) = -sinx

So, if f(x) = x⁴sinx, then f(-x) = -f(x). Thus, f(x) is an odd function. Therefore, the graph of y = f(x) is symmetric with respect to the origin.

So, ∫_{(-π/3 to π/3)} x⁴sinxdx = 0