Area functions Let A(x)

Question

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

Accepted Answer

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.

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>

Key Concepts

Definite Integral

Piecewise Functions

Area Under a Curve

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