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Evaluating a definite integral via rectangle and semicircle areas

Consider the piecewise function f(x)={2−1≤x<14−(x−3)2+21≤x≤5f(x) = \begin{cases} 2 & -1 \le x < 1 \\ \sqrt{4-(x-3)^2}+2 & 1 \le x \le 5 \end{cases}

The graph of f(x)f(x) on [−1,5][-1,5] consists of a horizontal segment followed by the top half of a circle shifted upward. Use the picture below to help identify the two regions.

Evaluating a definite integral via rectangle and semicircle areas graph

Evaluate ∫−15f(x) dx\displaystyle\int_{-1}^{5} f(x)\,dx by recognizing each piece as a standard geometric shape and using area formulas (no antiderivatives).

Hint

Split the integral at x=1x=1 using additivity. The first piece is a rectangle of height 2 over [−1,1][-1,1]. The second piece is a semicircle of radius 2 (centered at (3,2)(3,2)) sitting on top of a rectangle of height 2 over [1,5][1,5].

Step-by-step solution

By additivity, ∫−15f(x) dx=∫−112 dx+∫15(4−(x−3)2+2)dx\int_{-1}^{5} f(x)\,dx = \int_{-1}^{1} 2\,dx + \int_{1}^{5} \left(\sqrt{4-(x-3)^2}+2\right)dx. The first integral is the area of a rectangle with width 1−(−1)=21-(-1)=2 and height 22, giving 2×2=42\times 2 = 4.

For the second integral, split it into ∫154−(x−3)2 dx+∫152 dx\int_1^5 \sqrt{4-(x-3)^2}\,dx + \int_1^5 2\,dx. The first part is the area of the upper half of a circle of radius 22 (since 4−(x−3)2\sqrt{4-(x-3)^2} traces a semicircle centered at x=3x=3 from x=1x=1 to x=5x=5), which equals 12π(2)2=2π\frac{1}{2}\pi(2)^2 = 2\pi. The second part is a rectangle of width 44 and height 22, giving 88.

Adding all the pieces: 4+2π+8=12+2π4 + 2\pi + 8 = 12 + 2\pi. So ∫−15f(x) dx=12+2π\int_{-1}^{5} f(x)\,dx = 12 + 2\pi.

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