Question

1. for the function ( f ) graphed below evaluate the definite integrals

a) ( int_{-5}^{0} f(x) d x=)
b) ( int_{0}^{4} f(x) d x=)
c) ( int_{-6}^{6} f(x) d x=)

Explanation:

Step1: Calculate the area for part a)

The integral \(\int_{-5}^{0}f(x)dx\) is the area of a trapezoid and a triangle.
The trapezoid has bases \(b_1 = 3\) (from \(x=-5\) to \(x = - 3\)) and \(b_2=1\) (at \(x = 0\)) and height \(h = 2\). The area of the trapezoid \(A_{1}=\frac{(b_1 + b_2)h}{2}=\frac{(3 + 1)\times2}{2}=4\).
The triangle has base \(b = 3\) (from \(x=-3\) to \(x = 0\)) and height \(h = 3\). The area of the triangle \(A_{2}=\frac{1}{2}\times3\times3=\frac{9}{2}\). But since the part from \(x=-3\) to \(x = 0\) is below the \(x\)-axis, its contribution is negative.
\(\int_{-5}^{0}f(x)dx=3\times2-\frac{1}{2}\times3\times3=6 - 4.5 = 1.5\)

Step2: Calculate the area for part b)

The integral \(\int_{0}^{4}f(x)dx\) is the area of a triangle.
The base \(b = 4\) and height \(h=- 2\) (below the \(x\)-axis for \(0\leq x\leq2\) and above for \(2\leq x\leq4\)). The area of the two - triangle combination:
The area of the triangle from \(x = 0\) to \(x=2\): \(A_{3}=\frac{1}{2}\times2\times2 = 2\) (negative). The area of the triangle from \(x = 2\) to \(x=4\): \(A_{4}=\frac{1}{2}\times2\times2=2\) (positive). \(\int_{0}^{4}f(x)dx=- 2+2=0\)

Step3: Calculate the area for part c)

First, \(\int_{-6}^{-5}f(x)dx\): It's a rectangle with length \(l = 1\) and width \(w = 3\), so \(A_{5}=3\times1 = 3\)
We already know \(\int_{-5}^{0}f(x)dx = 1.5\) and \(\int_{0}^{4}f(x)dx=0\)
For \(\int_{4}^{6}f(x)dx\), it's a triangle with base \(b = 2\) and height \(h = 1\), \(A_{6}=\frac{1}{2}\times2\times1=1\)
\(\int_{-6}^{6}f(x)dx=\int_{-6}^{-5}f(x)dx+\int_{-5}^{0}f(x)dx+\int_{0}^{4}f(x)dx+\int_{4}^{6}f(x)dx\)
\(\int_{-6}^{6}f(x)dx=3 + 1.5+0 + 1=5.5\)

Answer:

a) \(1.5\)
b) \(0\)
c) \(5.5\)