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Chapter 4: Problem 7

Evaluate the definite integral of the algebraic function. Use a graphing utility to verify your result. $$ \int_{-1}^{0}(x-2) d x $$

Short Answer

Expert verified
The value of the definite integral is -1.

Step by step solution

01

Find the antiderivative

Find an antiderivative \(F(x)\) of \(f(x) = x - 2\). An antiderivative of \(x\) is \(\frac{1}{2}x^2\) and an antiderivative of \(-2\) is \(-2x\). So, \(F(x) = \frac{1}{2}x^2 - 2x\).
02

Evaluate the antiderivative at the limits of integration

Evaluate \(F(x)\) at 0 and -1. Given \(F(x) = \frac{1}{2}x^2 - 2x\), we find \(F(0) = \frac{1}{2}(0)^2 - 2(0) = 0\) and \(F(-1) = \frac{1}{2}(-1)^2 - 2(-1) = -1 + 2 = 1\).
03

Calculate the definite integral

Compute \(F(b) - F(a)\) to get the value of the definite integral. In this case, that is equal to \(F(0) - F(-1) = 0 - 1 = -1\).

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Most popular questions from this chapter

In Exercises \(69-74\), find the indefinite integral using the formulas of Theorem 4.24 \(\int \frac{1}{\sqrt{1+e^{2 x}}} d x\)

In Exercises \(79-84,\) find \(F^{\prime}(x)\). $$ F(x)=\int_{x}^{x+2}(4 t+1) d t $$

Find the limit. \(\lim _{x \rightarrow 0} \frac{\sinh x}{x}\)

(a) integrate to find \(F\) as a function of \(x\) and (b) demonstrate the Second Fundamental Theorem of Calculus by differentiating the result in part (a). $$ F(x)=\int_{1}^{x} \frac{1}{t} d t $$

Verify the differentiation formula. \(\frac{d}{d x}\left[\operatorname{sech}^{-1} x\right]=\frac{-1}{x \sqrt{1-x^{2}}}\)
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