Question

evaluate the following integrals by interpreting them in terms of areas:
(a) $$ \int _ { 0 } ^ { 2 } f ( x ) d x = $$
(b) $$ \int _ { 0 } ^ { 5 } f ( x ) d x = $$
(c) $$ \int _ { 5 } ^ { 7 } f ( x ) d x = $$
(d) $$ \int _ { 0 } ^ { 9 } f ( x ) d x = $$

Explanation:

Step1: Recall the integral - area relationship

The definite integral \(\int_{a}^{b}f(x)dx\) represents the net - signed area between the curve \(y = f(x)\), the \(x\) - axis, and the lines \(x=a\) and \(x = b\). Area above the \(x\) - axis is positive and area below the \(x\) - axis is negative. Assume each square has an area of \(1\) (since the grid is unit - based).

Step2: Calculate \(\int_{0}^{2}f(x)dx\)

The region from \(x = 0\) to \(x=2\) is a trapezoid. The formula for the area of a trapezoid is \(A=\frac{(a + b)h}{2}\), where \(a\) and \(b\) are the parallel sides and \(h\) is the height. Here, \(a = 1\), \(b = 3\), and \(h = 2\). So \(A=\frac{(1 + 3)\times2}{2}=4\).

Step3: Calculate \(\int_{0}^{5}f(x)dx\)

The region from \(x = 0\) to \(x = 2\) is a trapezoid (area \(A_1 = 4\)) and the region from \(x=2\) to \(x = 3\) is a rectangle (area \(A_2=3\times1 = 3\)) and the region from \(x = 3\) to \(x=5\) is a triangle. The formula for the area of a triangle is \(A=\frac{1}{2}bh\), where \(b = 2\) and \(h = 3\). So \(A_3=\frac{1}{2}\times2\times3 = 3\). Then \(\int_{0}^{5}f(x)dx=4 + 3+3=10\).

Step4: Calculate \(\int_{5}^{7}f(x)dx\)

The region from \(x = 5\) to \(x=7\) is a triangle below the \(x\) - axis (area is negative). Using \(A=\frac{1}{2}bh\), with \(b = 2\) and \(h = 2\), the area \(A=-\frac{1}{2}\times2\times2=- 2\).

Step5: Calculate \(\int_{0}^{9}f(x)dx\)

\(\int_{0}^{9}f(x)dx=\int_{0}^{5}f(x)dx+\int_{5}^{7}f(x)dx+\int_{7}^{9}f(x)dx\). We know \(\int_{0}^{5}f(x)dx = 10\), \(\int_{5}^{7}f(x)dx=-2\). The region from \(x = 7\) to \(x = 9\) is a triangle with \(b = 2\) and \(h = 1\), area \(A=\frac{1}{2}\times2\times1 = 1\). So \(\int_{0}^{9}f(x)dx=10-2 + 1=9\).

Answer:

(a) \(4\)
(b) \(10\)
(c) \(-2\)
(d) \(9\)