Calculus · Integrals

How to Evaluate the Definite Integral of a Sum

Problem

∫02(3x2+2x) dx\int_0^2 (3x^2 + 2x)\,dx

Answer

1212

A sum inside an integral splits into separate integrals. Integrate each term, then evaluate the result at both limits.

Step-by-step solution

  1. Find the antiderivative term by term.

    ∫(3x2+2x) dx=x3+x2\int (3x^2 + 2x)\,dx = x^3 + x^2

    Why: Integrate each term with the power rule.

  2. Write the limits on the bracket.

    [x3+x2]02\left[x^3 + x^2\right]_0^2

    Why: The limits stay on the bracket until both ends are evaluated.

  3. Evaluate at the top and bottom limits.

    (23+22)−(03+02)=12−0(2^3 + 2^2) - (0^3 + 0^2) = 12 - 0

    Why: Substitute the top limit, then the bottom limit, and subtract.

  4. Subtract.

    12−0=1212 - 0 = 12

    Why: The result is the signed area under the graph.

Common mistakes

  • Integrating the sum as if it were a product.
  • Subtracting bottom minus top.
  • Leaving + C in a definite integral. The limits remove the constant.

Practice problems

Use the same method. Work each problem on paper, then open the answer to check.

  1. ∫01(2x+1) dx\int_0^1 (2x + 1)\,dx

    Show answer

    22

  2. ∫02(x2+x) dx\int_0^2 (x^2 + x)\,dx

    Show answer

    143\frac{14}{3}

  3. ∫12(3x2+2x) dx\int_1^2 (3x^2 + 2x)\,dx

    Show answer

    1010

  4. ∫01(4x3+2x) dx\int_0^1 (4x^3 + 2x)\,dx

    Show answer

    22

  5. ∫02(6x2+2) dx\int_0^2 (6x^2 + 2)\,dx

    Show answer

    2020