Part 2 of the fundamental Theorem of Calculus

Question

Calculate the derivative

 

d/dx ∫x−1(4^t5−t)^22 dt =

 

using Part 2 of the Fundamental Theorem of Calculus.

 

d/dx ∫x−1(4t^5−t)^22 dt =

 

 __________

Answer 1

d/dx(∫_x^-1(4t⁵ - t)²²dt = - (4x⁵ - x)²²;

We get sign minus because x is lower limit

Answer 2

Hi Joaquin,

It looks like your problem is to calculate: d/dx { ∫_x⁻¹ (4^t5−t)^22 dt }, with integration limits x and -1.

The Fundamental Theorem of Calculus deals with integrals of the form ∫_a^x f(t) dt. We can put your integral into this form by multiplying by -1, which flips the integration limits:

d/dx { ∫_x⁻¹ (4^t5−t)^22 dt } = d/dx { ∫₋₁^x -1*(4^t5−t)^22 dt }

We now have an integral with the correct form, with a=-1 and f(t) = -1*(4^t5−t)^22.

The key idea is to realize that if you did actually evaluate this integral, then you would be left with a function of x, as in: F(x) = ∫₋₁^x -1*(4^t5−t)^22 dt .

So the question is asking you to compute d/dx { F(x) } . The Fundamental Theorem of Calculus prescribes that d/dx { F(x) } = f(x). In other words, replace t with x everywhere in f(t) above.

For more info, check out:

https://mathworld.wolfram.com/FundamentalTheoremsofCalculus.html

Hope this helps!