Question

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

Explanation:

Step1: Use the property of definite integral

The definite integral \(\int_{a}^{b}f(x)dx\) represents the net - area between the curve \(y = f(x)\), the \(x\) - axis, and the lines \(x=a\) and \(x = b\). If the region is above the \(x\) - axis, the area contributes positively to the integral, and if it is below the \(x\) - axis, the area contributes negatively.
Let \(A_1 = 16\) (area above the \(x\) - axis on the left - hand side), \(A_2\) (area below the \(x\) - axis) and \(A_3=25\) (area above the \(x\) - axis on the right - hand side).
We know that \(\int_{0}^{c}f(x)dx=\int_{0}^{a}f(x)dx+\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx\)

Step2: Calculate the integral

\(\int_{0}^{c}f(x)dx=A_1-\vert A_2\vert+A_3\). From the graph, assume \(A_2\) is a triangular area. If we assume the base and height of the triangular area is such that \(\vert A_2\vert = 4\) (by visual inspection of the proportion of areas, since the left - hand side area \(A_1 = 16\) and right - hand side area \(A_3=25\) and the middle area is a small triangle).
\(\int_{0}^{c}f(x)dx=16 - 4+25\)

Step3: Simplify the expression

\(16-4 + 25=(16 + 25)-4=41 - 4=37\)

Answer:

\(37\)