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Introduction to Integrals · Ages 17–18

Definite Integrals and Area Under a Curve Practice

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Designed for 12th grade calculus students. Calculate the exact area bounded by the given curve, the x-axis, and the specified limits.

  1. Find the area under f(x) = 4 from x = 1 to x = 5.

  2. Find the area under f(x) = 2x from x = 0 to x = 3.

  3. Find the area under f(x) = 3x^2 from x = 1 to x = 2.

  4. Find the area under f(x) = x^2 + 1 from x = 0 to x = 3.

  5. Find the area under f(x) = 6x^2 - 2x from x = 1 to x = 3.

  6. Find the area enclosed between the curve f(x) = 4 - x^2 and the x-axis.

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Definite Integrals and Area Under a Curve Practice

Answer key

  1. Find the area under f(x) = 4 from x = 1 to x = 5.

    16. (Antiderivative F(x) = 4x. F(5) - F(1) = 20 - 4 = 16).

  2. Find the area under f(x) = 2x from x = 0 to x = 3.

    9. (Antiderivative F(x) = x^2. F(3) - F(0) = 9 - 0 = 9).

  3. Find the area under f(x) = 3x^2 from x = 1 to x = 2.

    7. (Antiderivative F(x) = x^3. F(2) - F(1) = 8 - 1 = 7).

  4. Find the area under f(x) = x^2 + 1 from x = 0 to x = 3.

    12. (Antiderivative F(x) = (x^3)/3 + x. F(3) - F(0) = (9 + 3) - 0 = 12).

  5. Find the area under f(x) = 6x^2 - 2x from x = 1 to x = 3.

    44. (Antiderivative F(x) = 2x^3 - x^2. F(3) = 2(27) - 9 = 45; F(1) = 2(1) - 1 = 1; 45 - 1 = 44).

  6. Find the area enclosed between the curve f(x) = 4 - x^2 and the x-axis.

    32/3 (or 10 and 2/3). (Roots at x = -2 and x = 2. Antiderivative F(x) = 4x - (x^3)/3. F(2) - F(-2) = [8 - 8/3] - [-8 + 8/3] = 16/3 - (-16/3) = 32/3).