Area functions Let A(x) be the area of the region bounded by the t -axis and the graph of y=ƒ(t) from t=0 to t=x. Consider the following functions and graphs.

a. Find A(2) .

ƒ(t) = {-2t+8 if t ≤ 3 ; 2 if t >3 <IMAGE>

Video transcript

Welcome back everyone to another video. Determine the area D X enclosed by the x axis and the function Y equals JX from X equals 0 to X equals 2, given that JX is a piecewise function such that it is equal to x + 11 when X is less than or equal to 6 and 5 when X is greater than 6. So essentially we want to identify the area between X equals 0 and X equals 2. We are starting with the origin and then we can simply label X equals 2 on the graph. And essentially we're looking at the area below the line, right, because we have a line and a constant function. So we're looking at the area below the line. And above the x axis. We can recognize that this shape is a trapezoid, right? Essentially, if we draw two vertical lines, we will be able to see that. So let's go ahead and do that. And now if this is a trapezoid, well, essentially we can use the formula that allows us to identify the area of a trapezoid, which is simply A + B divided by 2 multiplied by H. A and B, those are the basis, and H is the height. We can immediately see that our height is 2. Right? Essentially H is equal to delta X. Which is X2 minus X1, and in this case, that's simply 2 minus 0, which is 2. So we want to identify the length of the base. Let's begin with A. Now for a we simply want to evaluate the function at 0.0, so we want to estimate G at 0. We're taking the line itself, right, because we're looking at the value which is less than or equal to 6. So J of 0 would be negative 0 + 11, which is simply 11, and that's the value of A. Now let's identify the length of B which is. J at 0.2. Substituting 2 into the function, we get 2 + 11, which is 9, and that's our B value. So we have the length of each base, we have the height, and this means that we can evaluate the area of the trapezoid, and that would be equal to. J at 0 plus J at 2 divided by 2. Multiplied by delta X, so we have simply rewritten it. And now we can say that A is equal to. 11 + 9 divided by 2. Multiplied by 2. We can see that this is equal to 20, right? Because we have a division by 2 and then we are multiplying by 2. So we simply have 20. And that's our final answer. The area is equal to 20. Thank you for watching.