1) You are given the four points in the plane A = (-2, 2), B = (1, -3), C = (6, 1), and D = (11,...


1) You are given the four points in the plane A = (-2, 2), B = (1, -3), C = (6, 1), and D = (11, -2).

The graph of the function f(x) consists of the three line segments AB, BC and CD.

Find the integral {eq}\int_{-2}^{11}f(x)dx {/eq} by interpreting the integral in terms of sums and/or differences of areas of elementary figures.

{eq}\int_{-2}^{11}f(x)dx= {/eq}

Integral as an Area

When we integrate a function over an interval, we are finding the area enclosed by the function and the x-axis between the endpoints of that interval. Therefore, if we can express this area as a known shape, such as a triangle, rectangle, circle, or other easy shape, we can find the area that way.

Answer and Explanation:

The function specified contains three line segments. The integral of a function represents the area between a function and the x-axis, so if we can find a way to express this area using known shapes, we can find the value of this integral. It turns out that the area between a straight line and the x-axis can indeed be thought of as a quadrilateral. If this is a horizontal line, this quadrilateral is a rectangle. Otherwise, it is a trapezoid. Since this function consists of three line segments, the area between it and the x-axis can be calculated using three trapezoids.

First, recall that the area of a trapezoid can be found as {eq}A = \frac{1}{2} (b_1+b_2)h {/eq}. Instead of having two bases, we have two heights, so we'll actually calculate this a little differently from how this area is typically defined {eq}A = \frac{1}{2} (h_1+h_2)\Delta x {/eq}. This means that our three areas is as follows.

{eq}\begin{align*} A_1 &= \frac{1}{2}(2+(-3))(1-(-2)) &= \frac{1}{2}(-1)(3)\\ &= -\frac{3}{2}\\ A_2 &= \frac{1}{2}(-3+1)(6-1)\\ &= \frac{1}{2}(-2)(5)\\ &= -5\\ A_3 &= \frac{1}{2}(1+(-2))(11-6)\\ &= \frac{1}{2}(-1)(5)\\ &= -\frac{5}{2} \end{align*} {/eq}

The integral is the sum of these three areas.

{eq}\begin{align*} \int_{-2}^{11} f(x) dx &= A_1 + A_2 + A_3\\ &= -1.5 - 5 - 2.5\\ &= -9 \end{align*} {/eq}

Learn more about this topic:

Calculating Integrals of Simple Shapes

from Math 104: Calculus

Chapter 13 / Lesson 1

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