Question

calculate the definite integral by referring to the figure with the indicated areas. area ( a = 1.322 ) area ( c = 5.459 ) area ( b = 2.206 ) area ( d = 1.792 ) ( int_{b}^{0} f(x) d x= )

Explanation:

Step1: Recall the property of definite integral and area relationship

The definite integral \(\int_{b}^{0}f(x)dx=-\int_{0}^{b}f(x)dx\). And \(\int_{0}^{b}f(x)dx\) is related to the areas. For the integral from \(b\) to \(c\) (area \(B\) is below the \(x -\)axis, so \(\int_{b}^{c}f(x)dx=- \text{Area of }B\)), and \(\int_{c}^{0}f(x)dx=\text{Area of }C\) (since it is above the \(x -\)axis).

Step2: Use the integral property \(\int_{b}^{0}f(x)dx=\int_{b}^{c}f(x)dx+\int_{c}^{0}f(x)dx\)

We know that \(\int_{b}^{c}f(x)dx=-\text{Area of }B\) (because the function \(y = f(x)\) is below the \(x -\)axis on \((b,c)\)) and \(\int_{c}^{0}f(x)dx=\text{Area of }C\) (because the function \(y = f(x)\) is above the \(x -\)axis on \((c,0)\)).
Substitute the values: \(\int_{b}^{0}f(x)dx=-\text{Area of }B+\text{Area of }C\).
Given \(\text{Area of }B = 2.206\) and \(\text{Area of }C=5.459\).

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Answer:

\(3.253\)