# Video: APCALC03AB-P1A-Q17-670182046420

The graph of a piecewise linear function π, for 0 β€ π₯ β€ 9, is shown. What is the value of β«_(0)^(9) π(π₯) dπ₯?

03:13

### Video Transcript

The graph of a piecewise linear function π, for zero is less than or equal to π₯ is less than or equal to nine, is shown. What is the value of the integral from zero to nine of π of π₯ with respect to π₯?

A definite integral, such as the one we have here, defines an area. Itβs the area of the region enclosed by the graph of the function weβre integrating β thatβs π of π₯ β the π₯ -axis, and two vertical lines at the limits of the integral. Those are the lines π₯ equals zero and π₯ equals nine in this case. This integral, therefore, corresponds to the area now shaded in orange on the given figure.

Now we donβt actually need to find the equation of our function π of π₯ in order to evaluate this integral, although we could look at finding the equations of each of its line segments if we wish. Instead, because π of π₯ is a piecewise linear function, this area is composed of two-dimensional shapes whose areas we can find using standard formulae.

We can split this area into two separate regions: region one and region two. Letβs consider region two first because this is just a triangle. The base of this triangle is just the difference between the π₯-coordinates at either of its ends. Thatβs nine minus three, which is equal to six units. And its perpendicular height can be found by looking at the vertical scale. We see that the height of this triangle is three units. Using the standard formula that the area of a triangle is equal to one-half multiplied by its base multiplied by its perpendicular height, we have that the area of region two is six multiplied by three over two, which is nine square units.

Now letβs consider area one. But we need to be very careful here because this area is below the π₯-axis. When we use integration to find an area below the π₯-axis, we get a negative result. The area itself isnβt negative, but the value of the integral used to find it is. To find the value of the contribution that area one makes to the integral then, we need to find the area of the two-dimensional shape, which is a trapezium, and then multiply it by negative one, as the area is below the π₯-axis.

The area of a trapezium or a trapezoid can be found by finding half the sum of its parallel sides, which are three and one units, and then multiplying this by the perpendicular distance between them, which is two units. So the area of this trapezoid is a half multiplied by three plus one multiplied by two. But its contribution to the integral is the negative of this value. This works out to be negative four.

The integral from zero to nine of our function π of π₯ with respect to π₯ then is equal to negative four plus nine, which is equal to five. Remember, the key point in this question is that areas below the π₯-axis are found to have a negative value when calculated using integration.