Question

Graph the function \(f(x)=\left(16+x^{2}\right)^{-3 / 2}\) and find the area of the region bounded by the curve and the \(x\) -axis on the interval [0,3]

Short answer

Answer: The area of the region is \(\frac{1}{60}\).

Step-by-step solution

Graph the function

To graph the function \(f(x)=\left(16+x^{2}\right)^{-\frac{3}{2}}\), we can plot a few points. The points should be chosen at reasonable intervals from the interval [0, 3]. For example, the values could be \(x=0,1,2,3\). Calculate the function values: - \(f(0)=\left(16+0\right)^{-\frac{3}{2}}=\frac{1}{4}\) - \(f(1)=\left(16+1\right)^{-\frac{3}{2}}=\frac{1}{27\sqrt{3}}\) - \(f(2)=\left(16+4\right)^{-\frac{3}{2}}=\frac{1}{216}\) - \(f(3)=\left(16+9\right)^{-\frac{3}{2}}=\frac{1}{125\sqrt{5}}\) By plotting the points \((0, \frac{1}{4})\), \((1, \frac{1}{27\sqrt{3}})\), \((2, \frac{1}{216})\), and \((3, \frac{1}{125\sqrt{5}})\), we obtain a decreasing curve that approaches the x-axis but never touches it.

Find the area under the curve

To find the area of the region bounded by the curve and the \(x\)-axis on the interval [0, 3], we need to compute the definite integral of the function over this interval. The integral is given by: \(\int_{0}^{3} \left(16+x^{2}\right)^{-\frac{3}{2}} dx\)

Perform a substitution (u-substitution)

To evaluate the integral, we can use a u-substitution method. Let \(u = 16 + x^2\). This means that \(du = 2x dx\). Now, we need to find the new bounds for the integral. When \(x=0\), \(u=16\). When \(x=3\), \(u=25\). So our integral becomes: \(\int_{16}^{25} u^{-\frac{3}{2}}\left(\frac{1}{2}\right) du\)

Evaluate the integral

Now we can evaluate the integral: \(\int_{16}^{25} u^{-\frac{3}{2}}\left(\frac{1}{2}\right) du = -\frac{1}{2}(\frac{2}{3})(\frac{1}{\sqrt{u}})\bigg\rvert_{16}^{25}\) Plug in the bounds: \(-\frac{1}{3}(\frac{1}{\sqrt{25}}-\frac{1}{\sqrt{16}})= -\frac{1}{3}( \frac{1}{5} - \frac{1}{4} )\)

Calculate the final result

Simplify the expression to find the area under the curve: \(-\frac{1}{3}( \frac{1}{5} - \frac{1}{4} ) = -\frac{1}{3}\times\frac{-1}{20}=\frac{1}{60}\) The area of the region bounded by the curve and the \(x\)-axis on the interval [0, 3] is \(\frac{1}{60}\).