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Chapter 11: Problem 7

Verify the statement by showing that the derivative of the right side is equal to the integrand of the left side. $$ \int(x-2)(x+2) d x=\frac{1}{3} x^{3}-4 x+C $$

Short Answer

Expert verified
By differentiating the given solution \(\frac{1}{3} x^{3}-4 x+C\), we obtain \(x^{2}-4\), which is the same as the original integrand when factored. Hence, the verification of the integral solution is confirmed.

Step by step solution

01

Understand the problem

To verify the solution \(\frac{1}{3} x^{3}-4 x+C\) of the given integral \(\int(x-2)(x+2) dx\), it is required to differentiate the solution and compare it with the original integrand \((x-2)(x+2)\).
02

Differentiate the solution

The derivative of \(\frac{1}{3} x^{3}-4 x+C\) with respect to \(x\) gives \(x^{2}-4\). This is calculated by applying the power rule, where the derivative of \(x^n\) is given by \(nx^{n-1}\), and the derivative of a constant term (\(C\)) is zero.
03

Simplify and compare with the original integrand

The integrand \((x-2)(x+2)\) simplifies to \(x^{2}-4\). Since the derivative of the solution and the simplified integrand are identical, this confirms that the given integral solution is correct.

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Most popular questions from this chapter

Find the area of the region. $$ \begin{aligned} &f(x)=3\left(x^{3}-x\right) \\ &g(x)=0 \end{aligned} $$

Use a graphing utility to graph the region bounded by the graphs of the functions. Write the definite integrals that represent the area of the region. (Hint: Multiple integrals may be necessary.) $$ f(x)=2 x, g(x)=4-2 x, h(x)=0 $$

Use the Midpoint Rule with \(n=4\) to approximate the area of the region bounded by the graph of \(f\) and the \(x\) -axis over the interval. Compare your result with the exact area. Sketch the region. $$ f(x)=4 x^{2} $$ $$ [0,2] $$

Use a symbolic integration utility to evaluate the definite integral. \(r^{6}\). $$ \int_{2}^{5}\left(\frac{1}{x^{2}}-\frac{1}{x^{3}}\right) d x $$

Use the Trapezoidal Rule with \(n=4\) to approximate the definite integral. $$ \int_{0}^{2} \frac{1}{x+1} d x $$
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