Evaluate the integral. ∫−22(x+3)2dx

Question

Evaluate the integral.  ∫−22(x+3)2dx

Robert Harris · Accepted Answer

Step 1  Given:  The integral   ∫−22(x+3)2dx  Step 2  Use substitution to integrate   Let  u=x+3  ⇒du=dx  When x = – 2 ,  u = – 2 + 3 = 1     When x = 2 , u = 2 + 3 = 5  Substituting u=x+3,  we get the integral   ∫15u2du  Step 3  To simplify further, use the power rule of integration  ∫xndx=xn+1(n+1)+C  ∫15u2du=[u33]15  =(533)−(13)  =1253−13  =125−13  =1243  Therefore,  ∫−22(x+3)2dx=1243

Susan Yang · Accepted Answer

Step 1: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:

\int_{a}^{b} f (x) d x = F (x) ∣_{a}^{b} = F (b) - F (a)

Step 2: In this case,

f (x) = {(x + 3)}^{2}

. Find its integral.

\frac{x^{3}}{3} + 3 x^{2} + 9 x ∣_{- 2}^{2}

Step 3: Since

F (x) ∣_{a}^{b} = F (b) - F (a)

, expand the above into F(2)−F(−2):

(\frac{2^{3}}{3} + 3 \times 2^{2} + 9 \times 2) - (\frac{{(- 2)}^{3}}{3} + 3 {(- 2)}^{2} + 9 \times - 2)

Step 4: Simplify.

\frac{124}{3}

Evaluate the integral. \int_{-2}^{2}(x+3)^{2}dx

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