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Post

Definite Integral of a Quadratic Function

Created by @nathanedwards

at November 1st 2023, 12:42:10 am.

AP_Calculus_AB-Questions

#integration

#antiderivative

#definite-integral

Question:

Find the definite integral of the function

f(x) = 3x^2 + 2x + 1

over the interval $[1, 4]$ .

Solution:

To find the definite integral of the function $f(x)$ over the interval $[a, b]$ , we need to evaluate the antiderivative of $f(x)$ and then substitute the upper and lower limits of integration.

Step 1: Evaluate the antiderivative of $f(x)$ :

F(x) = \int (3x^2 + 2x + 1) \, dx

Applying the power rule for integration,

\begin{align*} F(x) &= \int (3x^2 + 2x + 1) \, dx \\ &= \frac{3}{3}x^3 + \frac{2}{2}x^2 + x + C \\ &= x^3 + x^2 + x + C \end{align*}

where $C$ is the constant of integration.

Step 2: Evaluate the definite integral using the antiderivative obtained in Step 1:

The definite integral of $f(x)$ over the interval $[1, 4]$ is given by:

\begin{align*} \int_{1}^{4} (3x^2 + 2x + 1) \, dx &= F(4) - F(1) \\ &= (4^3 + 4^2 + 4) - (1^3 + 1^2 + 1) \\ &= (64 + 16 + 4) - (1 + 1 + 1) \\ &= (84) - (3) \\ &= 81 \end{align*}

Therefore, the definite integral of $f(x) = 3x^2 + 2x + 1$ over the interval $[1, 4]$ is equal to $81$ .