Area functions from graphs The graph of ...

Question

Area functions from graphs The graph of ƒ is given in the figure. A(𝓍) = ∫₀ˣ ƒ(t) dt and evaluate A(2), A(5), A(8), and A(12).

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Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says consider the graph of FFT, and below this we're given a graph that has FFT plotted on it. And FFT is a piecewise function. The first piece consists of a quarter circle with a radius of 3 units centered at the origin. The next piece is a linear piece and is decreasing from the point of 3.0 to the point of 52. The next piece is also a linear piece, and it's increasing from the point of 5 minus 2 to the point of 7.2, and the final piece is also a linear piece, and is decreasing from the point of 7.2 to the point of 10. Given that the area function is A X, which is equal to the interval from 0 to X of F of TDT, evaluate A of 5. So we need to evaluate AF 5, and using our definition or the area function, we see that AF 5 is going to be equal to the interval from 0 to 5 of F of TDT. So if we look at our function for FFT, on the interval between 0 and 5, we see that FFT consists of two of the pieces in our piece-wise function. It consists of a quarter circle with a radius of 3 units, centered at the origin, and we also have a decreasing linear piece that goes from the point of 3.0 to the point of 52. So we want to do is to break up our integral into two separate integrals, one for each of those pieces. That means A of 5 is going to be equal to. The integral, going from 0 to 3 of F of T DT. Plus the integral from 3 to 5 of F of T. DT. And what we want to do now is to evaluate each one of these integrals separately, and we're going to do this using geometry. So recall that the area beneath the curve. is what we get when we are integrating a function. So that means that the integral from 0 to 3 of F of T. DT Is this going to be equal to the area of that curve. And on the interval from 0 to 3, on a graph of FT, we have that quarter circle with a radius of 3 units centered at the origin. That means we just need to find the area of that quarter circle, which, if you recall, it's going to be equal to 1 divided by 4. Multiplied by pi, multiplied by the radius of that. Quarter circle, and in our case, we have a radius of 3 unit, so we'll have 3, and that quantity is going to be squared. And so when we simplify this expression, this is going to be equal to 9 pi divided by 4. The next interval that we need to evaluate using geometry is going to be the integral from 3 to 5 of F of T. DT And if we look at our graph here, the area underneath the curve is represented by a triangle. However, this area is below the T-axis. So that means that this area is actually negative since it falls below the T-axis. So, that means that our integral here is going to be equal to minus, and here we'll have the area of a triangle, which, if you recall, it's gonna be 1/2 multiplied by the base, and the base is the distance between T equal to 3 and T equal to 5, which in our case is gonna be 2. And that gets multiplied by the height of the triangle, which is the distance from Y equal to 0 to Y equal to -2, which is also 2. And so when we simplify this expression, we see that this integral is going to be equal to minus 2. So that means that AF 5. is going to be equal to. 9 pi divided by 4. -2, once we add these two integrals together. So that is the answer that we were looking for for this problem. So, I hope that this explanation has been useful, and I'll see you in the next video.

Area functions from graphs The graph of ƒ is given in the figure. A(𝓍) = ∫₀ˣ ƒ(t) dt and evaluate A(2), A(5), A(8), and A(12).

Key Concepts

Definite Integral

Area Under the Curve

Piecewise Functions

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