Question

Difficulty: EasyDefinite Integrals and Area Under Curves

What is the value of the definite integral 02(3x2+2)dx\int_{0}^{2} (3x^2 + 2) \, dx?

Answer: 12

Answer

12
To evaluate 02(3x2+2)dx\int_{0}^{2} (3x^2 + 2) \, dx, integrate 3x2+23x^2 + 2 to get x3+2xx^3 + 2x. Substituting the limits gives (23+2(2))(03+2(0))=(8+4)0=12(2^3 + 2(2)) - (0^3 + 2(0)) = (8 + 4) - 0 = 12.

Step-by-Step Solution

1
Integrate the function term by term
\int (3x^2 + 2) \, dx = x^3 + 2x
By the power rule of integration, \int 3x^2 \, dx = x^3 and \int 2 \, dx = 2x.
2
Apply the limits of integration from 0 to 2
[x^3 + 2x]_0^2 = (2^3 + 2(2)) - (0^3 + 2(0)) = 12 - 0 = 12
Substitute the upper limit 2 into the antiderivative and subtract the value obtained by substituting the lower limit 0.

Key Concept

Evaluation of Definite Integrals
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