Fundamental theorem of calculus part 1

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I’ve hit a brick wall in my math homework. There’s something obvious that I’ve forgotten, and I’m missing a connection in my brain somewhere.

“Let f(x) = 3- √x. Take the derivative of the function ∫^(2x\^3) ₄ f(t)dt using the Fundamental Theorem of Calculus, Part 1. If necessary, rewrite the integral expression first.”

It’s a multiple choice question and none of the answers include a “t”. Where did it go? What is up with the “t”? Why is it there? I swear I knew this like 3 days ago, but the knowledge left my brain as soon as I stopped looking at that problem, and I didn’t write it down. The textbook does not mention the “t” after stating the example problems, so I know whatever I’m doing to it is probably step 1.

In: 7

2 Answers

Anonymous 0 Comments

T is just a placeholder for time. Did you learn that integrals and double/triple integrals give you acceleration, velocity, etc.?

There is no T in the answer. Because you are plugging in the start/end points of the given definite integral.

The theorem just states that if the function is continuous in a range, then you can pick any start/end point in that range for the integral of that function and its derivative is the original function.

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> Let f(x) = 3- √x. Take the derivative of the function ∫2×3) ₄

I’ll take this as the start point is 4 and the end point is 2x/3.

So take the integral of 3- √x:

f(x) = 3 – x^(0.5)

∫f(x) = F(x) = 3x – (x^(1.5) /1.5 ) + c

Now do F(2x/3) – F(4)

[3(2x/3) – (2x/3)^(1.5) /1.5] – [3(4) – (4)^(1.5) /1.5]

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