Understanding the Fundamental Theorem of Calculus in Calculus II

[ardentmed]

Hey guys,

View attachment 2777I'm confused as to how these work. For instance, I'm assuming that the FTC must be used, where the formula used would be f(b)-f(a), correct?
If so, I got 4x^3 radical(2-x^4) -2 for the first one. The second one is 3x^2 (sin(x^3)) + sincosx * cosx.

Thanks in advance.

 

[MarkFL]

Let's take a look at the first one:

\(\displaystyle f(x)=\int_{-2}^{x^4}\sqrt{2-u}\,du\)

Let's define a function:

\(\displaystyle G(u)\) such that \(\displaystyle G'(u)=g(u)=\sqrt{2-u}\)

Now, by the anti-derivative form of the FTOC, we know:

\(\displaystyle f(x)=G\left(x^4\right)-G(-2)\)

Now, differentiate both sides with respect to $x$, making sure to use the chain rule where applicable. What do you get?

 

[ardentmed]

Let's take a look at the first one:

\(\displaystyle f(x)=\int_{-2}^{x^4}\sqrt{2-u}\,du\)

Let's define a function:

\(\displaystyle G(u)\) such that \(\displaystyle G'(u)=g(u)=\sqrt{2-u}\)

Now, by the anti-derivative form of the FTOC, we know:

\(\displaystyle f(x)=G\left(x^4\right)-G(-2)\)

Now, differentiate both sides with respect to $x$, making sure to use the chain rule where applicable. What do you get?

I used v=2-u as the substitution and ended up getting (-2/3)(2- x^4 )^(3/2) + (2/3)(8)

Thanks.

Edit: Also, I got cos(cos(x)) - cos( x^3 ) for the second one.

 

[MarkFL]

I used v=2-u as the substitution and ended up getting (-2/3)(2- x^4 )^(3/2) + (2/3)(8)

Thanks.

Edit: Also, I got cos(cos(x)) - cos( x^3 ) for the second one.

You shouldn't need to use any substitutions or actually integrate. Let's look at:

\(\displaystyle f(x)=G\left(x^4\right)-G(-2)\)

Now, differentiating with respect to $x$, we get:

\(\displaystyle f^{\prime}(x)=G^{\prime}\left(x^4\right)\left(x^4\right)^{\prime}-G^{\prime}(-2)(-2)^{\prime}\)

Now, using the fact that:

\(\displaystyle G^{\prime}(u)=g(u)=\sqrt{2-u}\)

we obtain (recalling the derivative of a constant is zero):

\(\displaystyle f^{\prime}(x)=\sqrt{2-x^4}\left(4x^3\right)-\sqrt{2-(-2)}(0)\)

Simplify:

\(\displaystyle f^{\prime}(x)=4x^3\sqrt{2-x^4}\)

Try the second problem in the same manner and post what you get. :D

 

[ardentmed]

You shouldn't need to use any substitutions or actually integrate. Let's look at:

\(\displaystyle f(x)=G\left(x^4\right)-G(-2)\)

Now, differentiating with respect to $x$, we get:

\(\displaystyle f^{\prime}(x)=G^{\prime}\left(x^4\right)\left(x^4\right)^{\prime}-G^{\prime}(-2)(-2)^{\prime}\)

Now, using the fact that:

\(\displaystyle G^{\prime}(u)=g(u)=\sqrt{2-u}\)

we obtain (recalling the derivative of a constant is zero):

\(\displaystyle f^{\prime}(x)=\sqrt{2-x^4}\left(4x^3\right)-\sqrt{2-(-2)}(0)\)

Simplify:

\(\displaystyle f^{\prime}(x)=4x^3\sqrt{2-x^4}\)

Try the second problem in the same manner and post what you get. :D

So 3x^2 (sin(x^3) + sin(cosx)sinx is what I computed for the second one after following your advice. Am I on the right track?

Thanks again.

 

[MarkFL]

So 3x^2 (sin(x^3) + sin(cosx)sinx is what I computed for the second one after following your advice. Am I on the right track?

Thanks again.

Good job! :D

I think you just have an extra bracketing symbol in there, but if you got:

\(\displaystyle g'(x)=3x^2\sin\left(x^3\right)+\sin(x)\sin\left(\cos(x)\right)\)

(which I think you did) then you are correct.