Question

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the graph of ( g ) is given below.
use the graph to evaluate each definite integral.
int_{-6}^{4} g(x) d x= int_{-4}^{4} g(x) d x= int_{-6}^{6} g(x) d x= int_{-5}^{-4} g(x) d x= int_{4}^{5} g(x) d x= int_{-4}^{5} g(x) d x= int_{-4}^{-2} g(x) d x= int_{-6}^{-5} g(x) d x=

Explanation:

Step1: Calculate the area of each geometric shape

The definite integral of a function \(y = g(x)\) from \(a\) to \(b\), \(\int_{a}^{b}g(x)dx\), is equal to the net - area between the curve \(y = g(x)\), 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.
We assume that each square on the grid has an area of \(1\) (since the distance between two consecutive grid lines is \(1\) unit).

Step2: Evaluate \(\int_{-6}^{4}g(x)dx\)

• For \(-6\leq x\leq - 4\): It 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, \(a = 3\), \(b = 2\), \(h=2\), and the area \(A_1=\frac{(3 + 2)\times2}{2}=5\) (negative since it is below the \(x\) - axis).
• For \(-4\leq x\leq - 2\): It is a rectangle with length \(2\) and width \(2\), area \(A_2=-2\times2=-4\).
• For \(-2\leq x\leq2\): It is a trapezoid. \(a = 2\), \(b = 1\), \(h = 4\), area \(A_3=\frac{(2 + 1)\times4}{2}=6\) (negative since it is below the \(x\) - axis).
• For \(2\leq x\leq4\): It is a trapezoid. \(a = 1\), \(b = 3\), \(h = 2\), area \(A_4=\frac{(1+3)\times2}{2}=4\) (positive since it is above the \(x\) - axis).

\(\int_{-6}^{4}g(x)dx=A_1 + A_2+A_3 + A_4=-5-4 - 6+4=-11\)

Step3: Evaluate \(\int_{-4}^{4}g(x)dx\)

• For \(-4\leq x\leq - 2\): Area \(A_2=-4\).
• For \(-2\leq x\leq2\): Area \(A_3=-6\).
• For \(2\leq x\leq4\): Area \(A_4 = 4\)

\(\int_{-4}^{4}g(x)dx=A_2+A_3 + A_4=-4-6 + 4=-6\)

Step4: Evaluate \(\int_{-6}^{6}g(x)dx\)

We already know \(\int_{-6}^{4}g(x)dx=-11\).

• For \(4\leq x\leq5\): It is a triangle with base \(1\) and height \(3\), area \(A_5=\frac{1\times3}{2}=1.5\) (positive).
• For \(5\leq x\leq6\): It is a triangle with base \(1\) and height \(5\), area \(A_6=-\frac{1\times5}{2}=-2.5\) (negative)

\(\int_{-6}^{6}g(x)dx=\int_{-6}^{4}g(x)dx+A_5+A_6=-11 + 1.5-2.5=-12\)

Step5: Evaluate \(\int_{-5}^{-4}g(x)dx\)

It is a trapezoid. \(a = 3\), \(b = 2\), \(h = 1\), area \(A=\frac{(3 + 2)\times1}{2}=2.5\) (negative)

\(\int_{-5}^{-4}g(x)dx=-2.5\)

Step6: Evaluate \(\int_{4}^{5}g(x)dx\)

It is a triangle with base \(1\) and height \(3\), area \(A=\frac{1\times3}{2}=1.5\) (positive)

\(\int_{4}^{5}g(x)dx = 1.5\)

Step7: Evaluate \(\int_{-4}^{5}g(x)dx\)

\(\int_{-4}^{5}g(x)dx=\int_{-4}^{4}g(x)dx+\int_{4}^{5}g(x)dx=-6 + 1.5=-4.5\)

Step8: Evaluate \(\int_{-4}^{-2}g(x)dx\)

It is a rectangle with length \(2\) and width \(2\), area \(A=-4\)

\(\int_{-4}^{-2}g(x)dx=-4\)

Step9: Evaluate \(\int_{-6}^{-5}g(x)dx\)

It is a trapezoid. \(a = 3\), \(b = 4\), \(h = 1\), area \(A=\frac{(3 + 4)\times1}{2}=3.5\) (negative)

\(\int_{-6}^{-5}g(x)dx=-3.5\)

Answer:

\(\int_{-6}^{4}g(x)dx=-11\), \(\int_{-4}^{4}g(x)dx=-6\), \(\int_{-6}^{6}g(x)dx=-12\), \(\int_{-5}^{-4}g(x)dx=-2.5\), \(\int_{4}^{5}g(x)dx = 1.5\), \(\int_{-4}^{5}g(x)dx=-4.5\), \(\int_{-4}^{-2}g(x)dx=-4\), \(\int_{-6}^{-5}g(x)dx=-3.5\)