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Chapter 4: Q. 50 (page 353)

Calculate the exact value of each definite integral in Exercises 47鈥�52 by using properties of definite integrals and the formulas in Theorem 4.13.
${鈭�}_{1}^{3} (x + 1)^{2} d x$

Short Answer

Expert verified

The exact value of definite integral is $\frac{56}{3}$ .

Step by step solution

01

Step 1. Given Information

We are given,

${鈭�}_{1}^{3} (x + 1)^{2} d x$

02

Step 2. Finding the Integral

The definite integral is given by,

${鈭�}_{1}^{3} (x + 1)^{2} d x = {鈭�}_{1}^{3} x^{2} + 2 x + 1 d x = {鈭�}_{1}^{3} x^{2} d x + 2 {鈭�}_{1}^{3} x + {鈭�}_{1}^{3} d x = \frac{1}{3} [3^{3} - 1] + 2 (\frac{1}{2}) [3^{2} - 1] + [3 - 1] = \frac{26}{3} + 8 + 2 = \frac{26 + 24 + 6}{3} = \frac{56}{3}$

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Most popular questions from this chapter

Use the Fundamental Theorem of Calculus to find the exact values of the given definite integrals. Use a graph to check your answer.

${鈭�}_{0}^{1} 鈥� \frac{1}{2 e^{x}} d x$

Given a simple proof that ${鈭�}_{k = 5}^{n} (a_{k} + b_{k}) = {鈭�}_{k = 5}^{n} a_{k} + {鈭�}_{k = 5}^{n} b_{k}$

Show that $F (x) = \frac{1}{2} e^{(x^{2})} (x^{2} - 1)$ is an anti-derivative of $f (x) = x^{3} e^{(x^{2})} .$

Write each expression in Exercises 41鈥�43 in one sigma notation (with some extra terms added to or subtracted from the sum, as necessary).

${鈭�}_{k = 1}^{40} \frac{1}{k} - {鈭�}_{k = 0}^{39} \frac{1}{k + 1}$

Consider the region between f and g on [0, 4] as in the

graph next at the left. (a) Draw the rectangles of the left-

sum approximation for the area of this region, with n = 8.

Then (b) express the area of the region with definite

integrals that do not involve absolute values.

See all solutions

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