Question

consider the graph of the function $f(x)$:
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 \(\int_{a}^{b}f(x)dx\) is equal to 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. Each square in the grid has an area of \(1\times1 = 1\) (assuming the side - length of each square is \(1\) unit).

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

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

We split the integral into two parts: \(\int_{0}^{2}f(x)dx+\int_{2}^{3}f(x)dx+\int_{3}^{5}f(x)dx\).

• \(\int_{0}^{2}f(x)dx = 4\) (from Step 2).
• \(\int_{2}^{3}f(x)dx\): The region from \(x = 2\) to \(x = 3\) is a rectangle with height \(3\) and width \(1\), so its area \(A_1=3\times1 = 3\).
• \(\int_{3}^{5}f(x)dx\): The region from \(x = 3\) to \(x = 5\) is a triangle. The formula for the area of a triangle is \(A=\frac{1}{2}bh\). Here, \(b = 2\) and \(h = 3\), so \(A_2=\frac{1}{2}\times2\times3=3\).

Then \(\int_{0}^{5}f(x)dx=4 + 3+3=10\).

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

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

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

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

Answer:

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