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Chapter 13: Chapter 13

Problem 1

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Evaluate the integrals. $$ \int_{-1}^{1}\left(x^{2}+2\right) d x $$

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The definite integral of the function \(f(x) = x^2 + 2\) over the interval \([-1, 1]\) is \(\frac{14}{3}\).
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TABLE OF CONTENTS :
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Step 1: Antiderivative of the function

To find the antiderivative of the function, we can apply the power rule, i.e., for any function of the form \(x^n\), its antiderivative is \(\frac{x^{n+1}}{n+1}\): $$ F(x) = \int (x^2 + 2)dx = \frac{x^3}{3} + 2x + C $$

Step 2: Apply the Fundamental Theorem of Calculus

Now, we will use the fundamental theorem of calculus to evaluate the definite integral: $$\int_{-1}^{1}\left(x^{2}+2\right) d x = F(1) - F(-1)$$

Step 3: Compute F(1) and F(-1)

Evaluate F(x) at x = 1 and x = -1: $$F(1) = \frac{1^3}{3} + (2 \times 1) = \frac{1}{3} + 2 = \frac{7}{3}$$ $$F(-1) = \frac{(-1)^3}{3} + (2 \times -1) = -\frac{1}{3} - 2 = -\frac{7}{3}$$

Step 4: Solve the definite integral

Now that we have calculated F(1) and F(-1), we can find the value of the definite integral: $$\int_{-1}^{1}\left(x^{2}+2\right) d x = F(1) - F(-1) = \frac{7}{3} - \left(-\frac{7}{3}\right) = \frac{7}{3} + \frac{7}{3} = \frac{14}{3} $$ The value of the definite integral is \(\frac{14}{3}\).

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