Question

note that the portion of the graph for $-3 < x < 0$ is the boundary of a circle centred at $(0,1)$. $\frac{9pi + 40}{4}$ $\frac{9pi + 48}{4}$ $\frac{9pi + 32}{4}$ $\frac{9pi + 56}{4}$ $\frac{9pi + 36}{4}$

Explanation:

Step1: Calculate the area of the triangle on the left

The base of the left - hand triangle is \(|- 3-(-4)| = 1\) and the height is \(1\). Using the formula for the area of a triangle \(A=\frac{1}{2}bh\), we have \(A_1=\frac{1}{2}\times1\times1=\frac{1}{2}\).

Step2: Calculate the area of the trapezoid

The formula for the area of a trapezoid is \(A=\frac{(a + b)h}{2}\). Here, \(a = 1\), \(b=2\) and \(h = 3\). So \(A_2=\frac{(1 + 2)\times3}{2}=\frac{9}{2}\).

Step3: Calculate the area of the quarter - circle

The equation of a circle is \((x - a)^2+(y - b)^2=r^2\). Since the circle is centered at \((0,1)\) and passes through \((-3,1)\), the radius \(r = 3\). The area of a full - circle is \(A=\pi r^2=\pi\times3^2 = 9\pi\). The area of the quarter - circle (for \(-3\lt x\lt0\)) is \(A_3=\frac{9\pi}{4}\).

Step4: Calculate the area of the triangle on the right

The base of the right - hand triangle is \(|4 - 3|=1\) and the height is \(2\). Using the formula for the area of a triangle \(A=\frac{1}{2}bh\), we have \(A_4=\frac{1}{2}\times1\times2 = 1\).

Step5: Calculate the area of the large triangle

The base of the large triangle (for \(x\gt3\)) is \(|4 - 3| = 1\) and the height is \(2\). Using the formula for the area of a triangle \(A=\frac{1}{2}bh\), we have \(A_5=\frac{1}{2}\times1\times2=1\).

Step6: Calculate the total area

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Answer:

\(\frac{9\pi + 48}{4}\)