Derivative with Part I of Fundemental Theorem of Calculus

[Jason76]

Find derivative using part I of the fundamental theorem of calculus.

\(\displaystyle g(x) = \int_{x}^{2} \cos \sqrt{9t} dt\)

It seems \(\displaystyle g'(x)\) would be \(\displaystyle \cos \sqrt{9x} dt\)

because looking at this other problem:

Find derivative using part I of the fundamental theorem of calculus.

\(\displaystyle g(x) = \int_{1}^{x} t^{4} dt \)

\(\displaystyle g'(x)= x^{4}\)

What is the difference in the two problems?

Note: Part I differs from part II of the theorem, in that in part II, you take the definite integral with x as one of the limits, then take the derivative of the answer.

 

[stapel]

Find derivative using part I of the fundamental theorem of calculus.

\(\displaystyle g(x) = \int_{x}^{2} \cos \sqrt{9t} dt\)

It seems \(\displaystyle g'(x)\) would be \(\displaystyle \cos \sqrt{9x} dt\)

because looking at this other problem:

Find derivative using part I of the fundamental theorem of calculus.

\(\displaystyle g(x) = \int_{1}^{x} t^{4} dt \)

\(\displaystyle g'(x)= x^{4}\)

What is the difference in the two problems?

Note: Part I differs from part II of the theorem, in that in part II, you take the definite integral with x as one of the limits, then take the derivative of the answer.

Do you have a rule for integrals for reversing the limits of integration? Something like:

. . . . .\(\displaystyle \displaystyle{\int_a^b\, f(t)\, dt \, =\, -\int_b^a\, f(t)\, dt}\)

What do you get if you apply that rule?

 

[Jason76]

Let's try it using part II of the theorem, since it would yield the same answer

First take the integral

\(\displaystyle \int \cos[(9t)^{1/2}] dt\)

\(\displaystyle \int \cos[3(t)^{1/2}] dt\)

 

[HallsofIvy]

Yes, you can do it that way. If you let \(\displaystyle u= 3t^{1/2}\), what is du?

But it is much easier to do it as you suggested with Stapel's correction: \(\displaystyle g(x)= -\int_2^x cos(\sqrt{9x}) dx\) so that \(\displaystyle \frac{dg}{dx}= \) what?

 

[Jason76]

Yes, you can do it that way. If you let \(\displaystyle u= 3t^{1/2}\), what is du?

But it is much easier to do it as you suggested with Stapel's correction: \(\displaystyle g(x)= -\int_2^x cos(\sqrt{9x}) dx\) so that \(\displaystyle \frac{dg}{dx}= \) what?

\(\displaystyle g(x)= -\int_2^x cos(\sqrt{9x}) dx\)

\(\displaystyle g(x)= -\int_2^x cos^{1/2}(u) dx\)

\(\displaystyle u = 3t^{1/2}\)

\(\displaystyle du = \dfrac{3}{2}t^{-1/2}\)

 

[pka]

\(\displaystyle g(x)= -\int_2^x cos(\sqrt{9x}) dx\)

\(\displaystyle g(x)= -\int_2^x cos^{1/2}(u) dx\)

\(\displaystyle u = 3t^{1/2}\)

\(\displaystyle du = \dfrac{3}{2}t^{-1/2}\)

What in the world are you doing? A u-substitution has nothing whatsoever to do with question.

If each of \(\displaystyle h~\&~g\) is a differentiable function function and \(\displaystyle \displaystyle F(x) = \int_{g(x)}^{h(x)} {f(t)dt} \)

then \(\displaystyle F'(x) = f \circ h(x) \cdot h'(x) - f \circ g(x) \cdot g'(x)\)

Notice that anti-differentiation has nothing to do with it.
It is a straight out differentiation problem.

 

[Jason76]

\(\displaystyle g(x) \int_{x}^{2} \cos(\sqrt{t}) dt\)

\(\displaystyle g'(x) = -\cos\sqrt{9x} \) is the answer. However, if the 2 and x were switched

\(\displaystyle g(x) \int_{2}^{x} \cos(\sqrt{t}) dt\)

then the answer would be

\(\displaystyle g'(x) = \cos\sqrt{9x} \)

 

[HallsofIvy]

\(\displaystyle g(x) \int_{x}^{2} \cos(\sqrt{t}) dt\)

\(\displaystyle g'(x) = -\cos\sqrt{9x} \) is the answer. However, if the 2 and x were switched

\(\displaystyle g(x) \int_{2}^{x} \cos(\sqrt{t}) dt\)

No! You have been told twice before in this thread that "\(\displaystyle \int_a^b f(t)dt= -\int_b^a f(t)dt\)"!

If \(\displaystyle g(x)= \int_x^2 \cos(\sqrt{t})dt\) then \(\displaystyle g(x)= -\int_2^x \cos(\sqrt{t}) dt\)

then the answer would be

\(\displaystyle g'(x) = \cos\sqrt{9x} \)