Area by geometry Use geometry to evaluate the...

Question

Area by geometry Use geometry to evaluate the following definite integrals, where the graph of ƒ is given in the figure.

(c) ∫₅⁷ ƒ(𝓍) d𝓍

Graph of a piecewise function showing y = f(x) with defined segments, plotted on a grid with x and y axes.

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says the graph of H of X is given below, compute the interval from 5 to 9 of H of XGX using geometry, and below the problem we're given a graph of H of X, and HF X consists of four linear line segments. We're also given 4 possible choices as our answers. For choice A, we have 10, for choice B, we have 0, for choice C, we have 5, and for choice D, we have minus 5. So we're asked to compute the integral from 5 to 9 of HXDX using geometry. And so, if you look at our graph, between X equal to 5 and X equal to 9, we see that H of X is just a single decreasing linear line segment. And it begins at the point of 5.5 and ends at the point of 9. minus 5, and it crosses the X axis at X equal to 7. So, we're going to do is use the point where our line here crosses the x-axis, to separate our interval into two separate intervals. So that means that the integral from 5 to 9 of H of X. DX is going to be equal to the interval from 5 to 7 of H of X. DX Plus, the interval from 7 to 9 of H of X. DX. So we're going to do is use geometry to evaluate each one of these integrals. So for the first integral, we have the integral from 5 to 7 of H of X. DX. I recall that this integral is going to be the area underneath our curve. So, if we look at our graph, we see that the area underneath the curve of H of X between X equal to 5 and X equal to 7 forms a triangle. That means this integral is just going to be equal to the area of that triangle. So, the integral is going to be equal to, and here we call your formula for the area of a triangle, then it's going to be one half, multiplied by the base of our triangle. And so the base of our triangle is the distance between X equal to 5 and X equal to 7, which in this case is going to be 2. And then that gets multiplied by the height of the triangle, and so the height of the triangle goes between Y equal to 0 and Y equal to 5, and that's going to be a distance of 5, so we'll multiply question here by 5. And when we evaluate our expression, we see that this integral is going to be equal to 5. Now we have to do something similar for the other integral. So we'll have the integral from 7 to 9 of H of X. DX Which is going to be equal to the area underneath the curve, and again, we look at a graph. That we have the area of a triangle. However, since our height here is going to be below the x-axis, that means our height is going to be negative. So, the area of this triangle is going to be 1 half, multiplied by its base, which is going to be the distance between X equal to 7 and X equal to 9, which is 2. And our height is going to be the distance between Y equal to 0 and Y equal to -5, which is going to be -5. And when we evaluate this expression, we see that it is going to be equal to -5. So, our integral from 5 to 9 of H of X. DX is going to be equal to 5 plus minus 5, and when we evaluate this question, this is going to be equal to 0. So that is the answer that we are looking for for this problem. And if we compare that to our answer choices, we see that the correct answer choice corresponds to answer choice B. So, I hope that this explanation has been useful, and I'll see you in the next video.

Area by geometry Use geometry to evaluate the following definite integrals, where the graph of ƒ is given in the figure.
(c) ∫₅⁷ ƒ(𝓍) d𝓍

Key Concepts

Definite Integral

Piecewise Function

Area Under the Curve

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