Question

Find the value of \(c\) for which the area of the figure bounded by the curves \(y=\frac{4}{x^{2}}, x=1\) and \(y=c\) is equal to \(\frac{9}{4}\).

Short answer

Answer: \(c=\frac{16}{7}\)

Step-by-step solution

Determine the integration limits

Since the figure is bounded by the curves \(y=\frac{4}{x^{2}}\), \(x=1\), and \(y=c\), we can find the limits of integration by finding where the two functions \(y=\frac{4}{x^{2}}\) and \(y=c\) intersect with the line \(x=1\). We should first find the value of \(y\) at \(x=1\) for the function \(y=\frac{4}{x^{2}}\). When \(x=1\): \(y=\frac{4}{1^2}=4\) So, the lower limit of integration is \(x=1\), and the upper limit of integration is the intersection between \(y=c\) and \(y=\frac{4}{x^2}\), which is at \(y=c\).

Integrate the function

Now we will integrate the function \(y=\frac{4}{x^{2}}\) with respect to \(x\) between the limits \(1\) and \(c\). \(\int_{1}^{c} \frac{4}{x^2} dx\) To integrate this, we use the power rule which states that \(\int x^n dx = \frac{x^{n+1}}{n+1}+C\), where \(n+1\neq0\). In our case, \(n=-2\), so the integral becomes: \(\int_{1}^{c}\frac{4}{x^2}dx = 4\int_{1}^{c}x^{-2}dx = 4\left(\frac{x^{-1}}{-1}\right)\bigg|_{1}^{c}\)

Evaluate the integral

Next, we'll evaluate the integral: \(4\left(\frac{x^{-1}}{-1}\right)\bigg|_{1}^{c} = -4\left(\frac{1}{c}-\frac{1}{1}\right) = -\frac{4}{c} + 4\) Since we know that the area of the figure is equal to \(\frac{9}{4}\), we can set the integral equal to this value: \(-\frac{4}{c} + 4 = \frac{9}{4}\)

Solve for c

Finally, we'll solve for \(c\): \(-\frac{4}{c} = \frac{9}{4} - 4\) \(-\frac{4}{c} = -\frac{7}{4}\) \(c = \frac{16}{7}\) Therefore, the value of \(c\) for which the area of the figure bounded by the curves \(y=\frac{4}{x^{2}}\), \(x=1\), and \(y=c\) is equal to \(\frac{9}{4}\) is \(c=\frac{16}{7}\).