Question

$\int_{-1}^{5} \sqrt{5 + 4x - x^2} \\, dx$

Explanation:

Step1: Complete the square inside the square root

First, we rewrite the quadratic expression inside the square root: \(5 + 4x - x^{2}=-(x^{2}-4x)+5\). Completing the square for \(x^{2}-4x\), we add and subtract \(4\) (since \((\frac{-4}{2})^{2}=4\)): \(x^{2}-4x=(x - 2)^{2}-4\). So, \(5 + 4x - x^{2}=-( (x - 2)^{2}-4)+5=9-(x - 2)^{2}\). Now the integral becomes \(\int_{-1}^{5}\sqrt{9-(x - 2)^{2}}dx\).

Step2: Recognize the integral as a circle equation

The expression \(\sqrt{9-(x - 2)^{2}}\) can be rewritten as \(y=\sqrt{9-(x - 2)^{2}}\), squaring both sides gives \(y^{2}=9-(x - 2)^{2}\), or \((x - 2)^{2}+y^{2}=9\) with \(y\geq0\). This is the upper half of a circle with center \((2,0)\) and radius \(r = 3\).

Step3: Determine the interval of integration

We are integrating from \(x=-1\) to \(x = 5\). Let's find the distance from the center \(x = 2\) to the limits. When \(x=-1\), the distance from \(x = 2\) is \(|2-(-1)|=3\), and when \(x = 5\), the distance from \(x = 2\) is \(|5 - 2|=3\). So the integral from \(x=-1\) to \(x = 5\) of \(\sqrt{9-(x - 2)^{2}}dx\) represents the area of the upper half of the circle (since \(y\geq0\)) from \(x=-1\) to \(x = 5\), which is actually the area of a semicircle? Wait, no. Wait, the circle has radius \(3\), and the interval from \(x=-1\) to \(x = 5\) is a length of \(5-(-1)=6\), which is equal to the diameter of the circle (since diameter \(d = 2r=6\)). So the region under the curve \(y=\sqrt{9-(x - 2)^{2}}\) from \(x=-1\) to \(x = 5\) is a semicircle? Wait, no. Wait, the center is at \(x = 2\), radius \(3\). So the leftmost point of the circle is \(x=2 - 3=-1\), and the rightmost point is \(x=2+3 = 5\). So the curve \(y=\sqrt{9-(x - 2)^{2}}\) from \(x=-1\) to \(x = 5\) is the upper half of the entire circle. Wait, the area of a full circle is \(\pi r^{2}\), so the area of the upper half (semicircle) is \(\frac{1}{2}\pi r^{2}\).

Step4: Calculate the area

Given \(r = 3\), the area of the semicircle is \(\frac{1}{2}\pi(3)^{2}=\frac{9}{2}\pi\). Wait, but let's confirm with the integral formula for \(\int\sqrt{r^{2}-(x - a)^{2}}dx\). The integral of \(\sqrt{r^{2}-(x - a)^{2}}dx\) from \(a - r\) to \(a + r\) is the area of the upper half - circle with radius \(r\), which is \(\frac{1}{2}\pi r^{2}\). Here, \(a = 2\), \(r = 3\), \(a - r=-1\), \(a + r = 5\), so the integral \(\int_{-1}^{5}\sqrt{9-(x - 2)^{2}}dx=\frac{1}{2}\pi(3)^{2}=\frac{9}{2}\pi\).

Answer:

\(\frac{9}{2}\pi\)