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Chapter 4: Problem 17

Evaluate the definite integral of the algebraic function. Use a graphing utility to verify your result. $$ \int_{0}^{3}\left|x^{2}-4\right| d x $$

Short Answer

Expert verified
The value of the definite integral from 0 to 3 of the absolute value of \(x^2-4\) is -2.

Step by step solution

01

Identify the interval of the integral

The interval of the integral is from 0 to 3. Note that the function inside the integral, \(x^2 - 4\), is negative for values of \(x < 2\) . Thus, the problem breaks down into two separate integrals over the intervals from \([0, 2)\), where the function is negative, and \([2, 3]\), where the function is positive.
02

Evaluation of first integral

The integral of the function over the interval \([0, 2)\) will require the modulus to change the sign of the function. The integral thus becomes: \[- \int_{0}^{2} (x^{2}-4) dx\]. Solving this integral will yield: \[- \left[\frac{x^{3}}{3} - 4x\right]_0^2 = -\frac{8}{3} + 8 \] .
03

Evaluation of second integral

The integral of the function over the intervals \([2, 3]\) stays the same. The integral thus becomes: \[\int_{2}^{3} (x^{2}-4) dx\]. Solving this integral will yield: \[\left[\frac{x^{3}}{3} - 4x\right]_2^3 = \frac{9}{3} - 12 + \frac{8}{3} - 8 \] .
04

Add the two calculated integrals

The final step is to add the results from steps 2 and 3 together to get the final result, the definite integral of the function from 0 to 3. The result is: \[-\frac{8}{3} + 8 + \frac{9}{3} - 12 + \frac{8}{3} - 8 = -2 \] .

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