Area functions and the Fundamental Theorem Consider the functi...

Question

Area functions and the Fundamental Theorem Consider the function

ƒ(t) = { t if ―2 ≤ t < 0

t²/2 if 0 ≤ t ≤ 2

and its graph shown below. Let F(𝓍) = ∫₋₁ˣ ƒ(t) dt and G(𝓍) = ∫₋₂ˣ ƒ(t) dt.

fig

(a) Evaluate F(―2) and F(2).

Accepted Answer

Hello. In this video we are going to be considering the piece-wise function given to us, whose graph is provided below. We want to find F of -4 and F of 2 if F of X is defined as the definite integral from -2 to X of little F of T D T. We are going to start this problem by solving for F of -4. Now, the capital F function is defined as this definite integral, whose upper bound is X. So if we are plugging -4 in for X, then we are replacing this upper bound with -4. That means that we are going to rewrite the definite integral as the definite integral from -2 to -4 of little F of TDT. Now, we need to define the integran. We want to define the integran as the part of the piecewise function that is between the bounds of -2 and -4. Well, we know that T is defined between -4 and 0 if we are on the part of the function that is defined as T + 1. And so what that means is that for this bound of the definite integral, we are going to define our integrand as T + 1. So that is going to allow us to rewrite the integrand as T + 1 DT. Now what we need to do is we need to solve for the definite integral. First, we are going to flip our bounds as you want our bounds to be from the smallest boundary to the largest boundary. That is going to give us -1, multiplied by the definite integral from -4 to -2 of T + 1 DT. Then we are going to take the antiderivative in each term of the integrand, and that is going to give us -1, multiplied by the quantity, 1/2 T squared plus T, and this will be evaluated from -4 to -2. By the fundamental theorem of calculus, we will have the following. We will have -1 multiplied by 1/2. -2 squared. Plus -2. Minus the lower boundary, which is 1/2 4 squad. + -4. Now what we need to do is we need to simplify. In the first half, we have -2 squad, which is going to simplify to 4, and 4 multiplied by 2, will give us 2. Plus -2 will simplify to be -2, and we will have 2 minus 2, which simplifies to 0. Now, in the second half, we have -4 squad, which simplifies to 16, and 16 multiplied by 2, gives us 8. + -4 will give us -4 and 8 minus 4 was simplified to be 4. So what we are left with is -1 multiplied by 0 minus 4, which is going to simplify to be 4, and that is going to be the value of F of -4. Now we are going to repeat this process in order to solve for capital F of 2. Now, if we plug in 2, we are going to replace the upper boundary of the definite integral to be from -2 to 2. Of little F of T D T. Now, what we want to work with is we want to define the integran to be some part of the piece wise function that is between -2 and positive2. However, the integran or the piece size function is split at the value of 0. And so what this means is that in order to find this part of the function, we are going to have to split this integral into a sum of integrals. Now, again, the cutoff point of the piece size function is going to be zero, so we can rewrite F of 2 to be the integral from -2 to 0 of little F of TDT. Plus the integral from 0 to 2 of little F of TDT. And now we just need to define the integrans with respect to the bounds. Now, in the first integral, we have the bounds of -2 to 0. Now, that is going to be defined in the first part of the piecewise function, which is between -4 and 0. So, the first integral is going to have the same integrand as the first part of this problem. We are going to define the integrand to be T + 1. Now, what about the part of the function that is defined between 0 and 2? While 0 and 2 is defined on the piecewise function as T2 + 1. So the second integral is going to have an integrand of T2 + 1. And so now what we need to do is we need to simplify this sum of integrals. Now, in the first integral, if we were to take the antiderivative of each term, we will be left with one half T squared plus T, and this will be evaluated from -2 to 0. And if we take the antiderivative of the 2nd integral, we will have the quantity 1/3 multiplied by T cubed plus T, and this will be evaluated from 0 to 2. And so now what we need to do is we need to go ahead and apply the fundamental theorem of calculus part 2 to both of the parts of the sum. So, if we were to plug in 0 in the first part into this function, we will end up with 0. So we will have 0 minus -2 plugged into the antiderivative, which will be 12 multiplied by -2 2 plus -2. Then we are going to add the second part with fundamental theorem part 2. We will plug in our upper bound, which is going to give us 1/3, multiplied by 2 cubed. Plus 2 Then we will have to subtract the value of the lower bound, but if we plug in the lower bound, we end up with zero. Now, we have evaluated this before. We know that if we simplify 1/2 multiplied by -2 squad, we will end up with 2, and 2 + -2 is going to give us 0. So this whole first half of the sum is going to simplify to be 0. Next, what we are going to do is we are going to simplify the second half. Now, 2 cubed is going to simplify to be 8. And 8 multiplied by 1/3 will leave us with 8/3, and 8/3 plus 2 is going to simplify to be 14/3. So what we are left with is 0 minus 0. Plus 14/3 minus 0, which is going to simplify to be positive 14/3, and this is going to be the solution of F of 2. So just to recap, we plugged in the values of -4 and 2 into the capital F function and evaluated by using the fundamental theorem of calculus part 2. F F of -4 gave us 4, and F of 2 gave us 14/3. And with that being said, the solution to this problem is going to be A. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

Area functions and the Fundamental Theorem Consider the function
ƒ(t) = { t      if  ―2 ≤ t < 0
t²/2    if    0 ≤ t ≤ 2
and its graph shown below. Let F(𝓍) = ∫₋₁ˣ ƒ(t) dt and G(𝓍) = ∫₋₂ˣ ƒ(t) dt.


(a) Evaluate F(―2) and F(2).

Key Concepts

Definite Integral

Fundamental Theorem of Calculus

Piecewise Function

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Area functions and the Fundamental Theorem Consider the functionƒ(t) = { t if ―2 ≤ t < 0t²/2 if 0 ≤ t ≤ 2and its graph shown below. Let F(𝓍) = ∫₋₁ˣ ƒ(t) dt and G(𝓍) = ∫₋₂ˣ ƒ(t) dt. (a) Evaluate F(―2) and F(2).

Key Concepts

Definite Integral

Fundamental Theorem of Calculus

Piecewise Function

Watch next

Area functions and the Fundamental Theorem Consider the function
ƒ(t) = { t if ―2 ≤ t < 0
t²/2 if 0 ≤ t ≤ 2
and its graph shown below. Let F(𝓍) = ∫₋₁ˣ ƒ(t) dt and G(𝓍) = ∫₋₂ˣ ƒ(t) dt.

(a) Evaluate F(―2) and F(2).