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Chapter 4: Q. 64 (page 373)

Use the Fundamental Theorem of Calculus to find the exact
values of each of the definite integrals in Exercises $19 鈥� 64$ . Use
a graph to check your answer.
${鈭�}_{- 1}^{1} |e^{2 x} - 1| d x$

Short Answer

Expert verified

The value of integral is $- \frac{e^{2}}{2} + 1 + \frac{1 + 2 e^{2}}{2 e^{2}}$ and the plot is

Step by step solution

01

Step 1. Given information

An expression is given as ${鈭�}_{- 1}^{1} |e^{2 x} - 1| d x$

02

Step 2. Evaluating integral

Evaluate the integral as,

${鈭�}_{-}^{1} |e^{2 x} - 1| d x = {鈭�}_{- 1}^{0} - e^{2 x} + 1 d x + {鈭�}_{0}^{1} e^{2 x} - 1 d x = {[\frac{- e^{2 x}}{2}]}_{- 1}^{0} + [x]_{- 1}^{0} + {[\frac{e^{2 x}}{2}]}_{0}^{1} - [x]_{0}^{1} = [\frac{- e^{2 (0)}}{2} + \frac{e^{2 (- 1)}}{2}] + [0 + 1] - [\frac{e^{2}}{2} - \frac{e^{2 (0)}}{2}] - [1 + 0] = [- \frac{1}{2} + \frac{e^{- 2}}{2} + 1] + [- \frac{e^{2}}{2} + \frac{1}{2} + 1] = \frac{1}{2} + \frac{1}{2 e^{2}} + \frac{1}{2} - \frac{e^{2}}{2} = - \frac{e^{2}}{2} + 1 + \frac{1 + 2 e^{2}}{2 e^{2}}$

And the plot is

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Most popular questions from this chapter

Use integration formulas to solve each integral in Exercises 21鈥�62. You may have to use algebra, educated guess- and- check, and/or recognize an integrand as the result of a product, quotient, or chain rule calculation. Check each of your answers by differentiating. (Hint for Exercise 54: $\tan (x) = \frac{\sin (x)}{\cos (x)}$ ).

$鈭� (\sec^{2} (x) + \csc^{2} (x)) d x .$

Find the sum or quantity without completely expanding or calculating any sums.

Given ${鈭�}_{k = 0}^{25} k = 325$ and ${鈭�}_{k = 3}^{28} {(k - 3)}^{2} = 14, 910$ , find the value of ${鈭�}_{k = 3}^{25} (k^{2} - 5 k + 9)$ .

Fill in each of the blanks:

(a) $鈭� d x = x^{6} + C .$

(b) $x^{6}$ is an antiderivative of .

(c) The derivative of $x^{6}$ is .

Your calculator should be able to approximate the area between a graph and the x-axis. Determine how to do this on your particular calculator, and then, in Exercises 21鈥�26, use the method to approximate the signed area between the graph of each function f and the x-axis on the given interval [a, b].

$f (x) = 1 - 2^{x}, [a, b] = [- 3, 1]$

Describe an example that illustrates that ${鈭�}_{a}^{b} | f (x) | d x$ is not equal to $| {鈭�}_{a}^{b} f (x) d x |$ .

See all solutions

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