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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 definite integral of the function \(x - 2\) from -1 to 0 is 3.

Step by step solution

01

Application of the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if a function \(f\) is continuous over the interval \([a,b]\) and \(F\) is an antiderivative of \(f\) on \([a,b]\), then \(\int_a^b f(x) dx = F(b) - F(a)\). In this case, the function \(f(x)\) is a linear equation \(x-2\), whose antiderivative \(F(x)\) is \(\frac{x^2}{2} -2x\). Apply the theorem, we have \(\int_{-1}^0 (x - 2) dx = [\frac{x^2}{2} - 2x]_{-1}^0 = \frac{0^2}{2} - 2(0) - [\frac{-1^2}{2} - 2(-1)] = - (-3) = 3\)
02

Verification using a graphing tool

Draw the function \(f(x) = x -2\) and the area under the curve from -1 to 0 in a graphing tool. The area under the curve from -1 to 0 should be 3, which agrees with the result obtained in Step 1, thereby verifying the solution.

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