Question

the graph of ( f ) is shown below on the interval (-5, 3).

the area bounded between the graph of ( f ) and the ( x )-axis on (-5, -4) is ( \frac{10}{3} ), the area bounded between the graph of ( f ) and the ( x )-axis on (-4, 2) is 36, and the area bounded between the graph of ( f ) and the ( x )-axis on (2, 3) is ( \frac{10}{3} ). determine ( int_{-5}^{2} f(x) , dx ).

Explanation:

Step1: Analyze the sign of \( f(x) \) on each interval

On \([-5, -4]\), the graph of \( f(x) \) is above the \( x \)-axis (since the area is positive and the integral of a function above the \( x \)-axis is positive), so \( \int_{-5}^{-4} f(x) dx=\frac{10}{3} \). On \([-4, 2]\), the graph of \( f(x) \) is below the \( x \)-axis (the parabola opens upwards and between the roots \(-4\) and \(2\) it is below the \( x \)-axis), so the integral \( \int_{-4}^{2} f(x) dx=- 36\) (because the area between the graph and the \( x \)-axis is \(36\), but the function is negative there). On \([2, 3]\), the graph of \( f(x) \) is above the \( x \)-axis, so \( \int_{2}^{3} f(x) dx=\frac{10}{3} \), but we need \( \int_{-5}^{2} f(x) dx=\int_{-5}^{-4} f(x) dx+\int_{-4}^{2} f(x) dx\)

Step2: Calculate \( \int_{-5}^{2} f(x) dx \)

We know that \( \int_{-5}^{-4} f(x) dx = \frac{10}{3}\) and \( \int_{-4}^{2} f(x) dx=-36\) (because the function is below the \( x \)-axis, the integral is the negative of the area). So we sum these two integrals:
\( \int_{-5}^{2} f(x) dx=\int_{-5}^{-4} f(x) dx+\int_{-4}^{2} f(x) dx=\frac{10}{3}+(- 36)\)

First, convert \( 36\) to thirds: \( 36=\frac{108}{3}\)

Then \( \frac{10}{3}-\frac{108}{3}=\frac{10 - 108}{3}=\frac{- 98}{3}\)? Wait, no, wait. Wait, the interval \([-4,2]\): the area between the graph and the \( x \)-axis is \(36\), but since the function is below the \( x \)-axis, the integral \( \int_{-4}^{2}f(x)dx=- \text{Area}=-36\). And \( \int_{-5}^{-4}f(x)dx=\text{Area}=\frac{10}{3}\) (since above \( x \)-axis). Then \( \int_{-5}^{2}f(x)dx=\int_{-5}^{-4}f(x)dx+\int_{-4}^{2}f(x)dx=\frac{10}{3}-36\)

Calculate \( 36=\frac{108}{3}\), so \( \frac{10}{3}-\frac{108}{3}=\frac{10 - 108}{3}=\frac{-98}{3}\)? Wait, that can't be right. Wait, maybe I made a mistake in the sign of \( \int_{-4}^{2}f(x)dx \). Wait, the graph is a parabola opening upwards with roots at \( x = - 4\) and \( x=2\). So between \( x=-4\) and \( x = 2\), the function \( f(x)\leq0\), so the integral of \( f(x) \) from \( - 4\) to \( 2\) is equal to the negative of the area between the graph and the \( x \)-axis. The area is \(36\), so \( \int_{-4}^{2}f(x)dx=-36\). The integral from \( - 5\) to \( - 4\): the graph is above the \( x \)-axis, so the integral is equal to the area, which is \( \frac{10}{3}\). Then \( \int_{-5}^{2}f(x)dx=\frac{10}{3}+(-36)=\frac{10 - 108}{3}=\frac{-98}{3}\)? Wait, but let's check again. Wait, the problem says "the area bounded between the graph of \( f \) and the \( x \)-axis on \([-4,2]\) is \(36\)". For a function that is below the \( x \)-axis, \( \int_{a}^{b}f(x)dx=-\text{Area}\) (because \( f(x)\leq0\), so the integral is negative, and the area is the absolute value). So yes, \( \int_{-4}^{2}f(x)dx=-36\). And \( \int_{-5}^{-4}f(x)dx=\frac{10}{3}\) (since above \( x \)-axis, integral equals area). Then \( \int_{-5}^{2}f(x)dx=\int_{-5}^{-4}f(x)dx+\int_{-4}^{2}f(x)dx=\frac{10}{3}-36\)

Convert \( 36\) to fraction with denominator \(3\): \( 36=\frac{108}{3}\)

So \( \frac{10}{3}-\frac{108}{3}=\frac{10 - 108}{3}=\frac{-98}{3}\)? Wait, that seems odd, but let's follow the steps.

Wait, maybe I misread the interval. The problem is to find \( \int_{-5}^{2}f(x)dx\). The intervals are \([-5,-4]\) (area \( \frac{10}{3}\), function above \( x \)-axis, so integral \( \frac{10}{3}\)), \([-4,2]\) (area \(36\), function below \( x \)-axis, so integral \( - 36\)). So adding them: \( \frac{10}{3}-36=\frac{10 - 108}{3}=\frac{-98}{3}\)? Wait, but let's check the calculation again.

\( 36\times3 = 108\), so \( 36=\frac{108}{3}\). Then \( \frac{10}{3}-\frac{108}{3}=\frac{10 - 108}{3}=\frac{-98}{3}\approx - 32.67\)

Wait, but let's confirm the sign of the integral on \([-4,2]\). If the graph is a parabola opening upwards with roots at \( x=-4\) and \( x = 2\), then for \( x\in(-4,2)\), \( f(x)<0\) (since it's below the \( x \)-axis between the roots of a upward - opening parabola). So the integral of \( f(x) \) from \( - 4\) to \( 2\) is the integral of a negative function, so it should be negative, and its absolute value is the area (36). So \( \int_{-4}^{2}f(x)dx=-36\) is correct. And \( \int_{-5}^{-4}f(x)dx\) is the integral of a positive function (above \( x \)-axis), so it's equal to the area \( \frac{10}{3}\). Then the sum is \( \frac{10}{3}-36=\frac{10 - 108}{3}=\frac{-98}{3}\)

Answer:

\(\boxed{-\dfrac{98}{3}}\)