Consider the function
ƒ(t) = { t &...

Question

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.

(f) Find a constant C such that F(𝓍) = G(𝓍) + C .

Accepted Answer

Hello. In this video, we are given a piecewise function whose graph is provided below. We are going to let F of X equal to the definite integral from -2 X of little F of TDT, and we are going to allow capital G of X to equal to -3 X of F of TDT. We want to determine the constant C such that capital F of X is equal to capital G of X plus C. Now, if we are to take a look at the given definite integrals, notice that F of X has the bounds from -2 to X, and capital G of X has the bounds from -3 to X. Since G of X has the bigger values for the bounds, that means that we can rewrite G of X as a sum of definite integrals that contains F of X. So again, we know that G of X. is equal to the definite integral from -3 to X of little F of TDT. Now we can split this up into a sum of integrals if we were to cut the integral in half at some value between -3 and X. In this case here, let's go ahead and choose -2. So we can rewrite G of X to be the sum of the definite integral from -3 to -2. Of little F of TDT plus the definite integral from -2 to X of little F of TDT. Notice that the integral from negative to to X of little F of TDT matches the given capital F of X function. And so with that being said, we can further rewrite this to B G of X is equal to the definite integral from 3 to -2 of little F of TDT plus F of X. Now, the statement here says that we want to solve for C for the function F of X is equal to GX plus C. And so if we were to try to rewrite this by solving for F of X, we can rewrite this to be F of X. is equal to G of X minus the definite integral from 3 to -2 of F of TDT. And so what we need to do now is we need to go ahead and solve this integral, since this integral is representing the constant C that we want to solve for. So if we write this on the side, we know that C is equal to the integral from -3 to -2 of little F of TDT. Now the function that is given by the P size function that is defined between -3 and -2 is defined as T + 1. Since the value of T in this case is between -4 and 0, and so the integrand for our definite integral is going to be T + 1 DT. And so now what we need to do is we need to take the antiderivative of each of the terms. That is going to give us one half T squared plus T, and this is going to be evaluated from -3 to -2. And by using the fundamental theorem of calculus part two, this is going to give us the quantity 1/2 multiplied by -2 2 plus -2, which simplifies to -2. Minus the quantity, 1/2 multiplied by -3 squad, plus -3, which is -3. And now what we need to do is simplify. -2 2 simplifies to 4, and 4 multiplied by 2 gives us 2. So we have 2 minus 2, minus -3 squad, which is 9, multiplied by 1/2, which is 9/12, minus 3. 2 minus 2 gives us 0. And 9/1/2 minus 3 gives us 3 halves. And so this is going to simplify to be negative 3 halves. Now if we were to take this and plug this back into our earlier F of X statement, this is going to give us F of X is equal to G of X minus -3 divided by 2, and by simplifying, this is going to give us G of X plus 3 divided by 2. And so that means that the constant value that gives us the equation of the form F of X equal to G X plus C. is going to be Cs equal to 3 halves. And with that being said, the overall solution to this problem is going to be D. 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.

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.

(f) Find a constant C such that F(𝓍) = G(𝓍) + C .

Key Concepts

Piecewise Functions

Definite Integrals

Fundamental Theorem of Calculus

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