Question

the accompanying figure shows the area of regions bounded by the graph of \\( y = f(x) \\) and the x - axis. evaluate the following integral \\( \int_{0}^{c} f(x) dx \\) \\( \int_{0}^{c} f(x) dx = \square \\) (simplify your answer)

Explanation:

Step1: Use the property of definite integral

We know that \(\int_{a}^{b}f(x)dx\) represents the net - area between the curve \(y = f(x)\), the \(x\) - axis, and the lines \(x=a\) and \(x = b\). If the area above the \(x\) - axis is positive and the area below the \(x\) - axis is negative.
Let \(A_1=16\) (area above the \(x\) - axis) and \(A_2 = 12\) (area below the \(x\) - axis).
The formula for \(\int_{a}^{b}f(x)dx=\text{(Area above the }x-\text{axis)}-\text{(Area below the }x-\text{axis)}\)

Step2: Calculate the integral \(\int_{0}^{5}f(x)dx\)

For \(\int_{0}^{5}f(x)dx\), using the net - area formula \(\int_{0}^{5}f(x)dx=A_1 - A_2\)
Substitute \(A_1 = 16\) and \(A_2=12\) into the formula: \(\int_{0}^{5}f(x)dx=16-12\)

Answer:

\(4\)