1. ## antiderivative question

this was a question on a calc test.

Let F(x) be the antiderivative of f(x) = (x^2) cos(2x) and F(1) = 0
a) find F(x) and F(3)
b) find the critical values of F(x) in [0,pi]
c) find F''(x)
d) For the values you found in b), explain which (if any) are local maxima. Justify your answer.
e) Find the absolute maximum value of F(x).

now these questions are pretty straightforward and easy but the problem i have is that there is no graphing calculator allowed on this test. i answered a,b,c, and, d easily, but e) is giving me some trouble. i found the local max at 3pi/4 and the two endpoints 0 and pi i have to test as well by plugging back into F(x). now this test was before we learned integration by parts so we can't use that technique. so the only way to find F(x) was to use the fundamental theorem of calculus so that F(x) = integral (from 1 to x) of (t^2)cos(2t) dt and evaluating F(x) means evaluating the respective integrals. the problem i have is that i don't know how to evaluate these integrals without a graphing calculator. My TI-84 has a program for evaluating definite integrals but the scientific calculators provided for us has no such function. how are you supposed to do part e) using only a scientific calculator?

2. Originally Posted by oblixps
this was a question on a calc test.

Let F(x) be the antiderivative of f(x) = (x^2) cos(2x) and F(1) = 0
a) find F(x) and F(3)
b) find the critical values of F(x) in [0,pi]
c) find F''(x)
d) For the values you found in b), explain which (if any) are local maxima. Justify your answer.
e) Find the absolute maximum value of F(x).

now these questions are pretty straightforward and easy but the problem i have is that there is no graphing calculator allowed on this test. i answered a,b,c, and, d easily, but e) is giving me some trouble. i found the local max at 3pi/4 and the two endpoints 0 and pi i have to test as well by plugging back into F(x). now this test was before we learned integration by parts so we can't use that technique. so the only way to find F(x) was to use the fundamental theorem of calculus so that F(x) = integral (from 1 to x) of (t^2)cos(2t) dt and evaluating F(x) means evaluating the respective integrals. the problem i have is that i don't know how to evaluate these integrals without a graphing calculator. My TI-84 has a program for evaluating definite integrals but the scientific calculators provided for us has no such function. how are you supposed to do part e) using only a scientific calculator?
To find $\displaystyle F(x)$ you need to use Integration by Parts twice.

Remember that $\displaystyle \int{u\,dv} = uv - \int{v\,du}$.

$\displaystyle F(x) = \int{x^2\cos{2x}\,dx}$

Let $\displaystyle u = x^2$ so that $\displaystyle du = 2x$.

Let $\displaystyle dv = \cos{2x}$ so that $\displaystyle v = \frac{1}{2}\sin{2x}$.

Then $\displaystyle F(x) = \frac{1}{2}x^2\sin{2x} - \int{x\sin{2x}\,dx}$

Now let $\displaystyle u = x$ so that $\displaystyle du = 1$

Let $\displaystyle dv = \sin{2x}$ so that $\displaystyle v = -\frac{1}{2}\cos{2x}$.

Then $\displaystyle F(x) = \frac{1}{2}x^2\sin{2x} - \left[-\frac{1}{2}x\cos{2x} - \int{-\frac{1}{2}\cos{2x}\,dx}\right]$

$\displaystyle = \frac{1}{2}x^2\sin{2x} + \frac{1}{2}x\cos{2x} - \frac{1}{4}\sin{2x} + C$.

Now use your initial condition to find $\displaystyle C$, and then you should be able to find $\displaystyle F(3)$.

3. Out of interest, what did you find F(x) to be?

4. Originally Posted by pickslides
Out of interest, what did you find F(x) to be?
Me or the OP?

5. Originally Posted by Prove It
To find $\displaystyle F(x)$ you need to use Integration by Parts twice.

Remember that $\displaystyle \int{u\,dv} = uv - \int{v\,du}$.

$\displaystyle F(x) = \int{x^2\cos{2x}\,dx}$

Let $\displaystyle u = x^2$ so that $\displaystyle du = 2x$.

Let $\displaystyle dv = \cos{2x}$ so that $\displaystyle v = \frac{1}{2}\sin{2x}$.

Then $\displaystyle F(x) = \frac{1}{2}x^2\sin{2x} - \int{x\sin{2x}\,dx}$

Now let $\displaystyle u = x$ so that $\displaystyle du = 1$

Let $\displaystyle dv = \sin{2x}$ so that $\displaystyle v = -\frac{1}{2}\cos{2x}$.

Then $\displaystyle F(x) = \frac{1}{2}x^2\sin{2x} - \left[-\frac{1}{2}x\cos{2x} - \int{-\frac{1}{2}\cos{2x}\,dx}\right]$

$\displaystyle = \frac{1}{2}x^2\sin{2x} + \frac{1}{2}x\cos{2x} - \frac{1}{4}\sin{2x} + C$.

Now use your initial condition to find $\displaystyle C$, and then you should be able to find $\displaystyle F(3)$.

All for naught pal...he explicitly said they cant use I.B.P.

6. Originally Posted by vince
All for naught pal...he explicitly said they cant use I.B.P.
I should have read the question properly.

As an alternative to Integration by parts, there's the "Find the derivative and create a new integral equation" method.

$\displaystyle \frac{d}{dx}(x^2\cos{2x}) = 2x^2\sin{2x} + 2x\cos{2x}$.

So $\displaystyle \int{2x^2\sin{2x}\,dx} + \int{2x\cos{2x}\,dx} = x^2\cos{2x}$

Now to find these integrals, differentiate each of them...

$\displaystyle \frac{d}{dx}(2x^2\sin{2x}) = 4x^2\cos{2x} + 4x\sin{2x}$

So $\displaystyle \int{4x^2\cos{2x}\,dx} + \int{4x\sin{2x}\,dx} = 2x^2\sin{2x}$

Therefore $\displaystyle 4\int{x^2\cos{2x}\,dx} = 2x^2\sin{2x} - 4\int{x\sin{2x}\,dx}$

$\displaystyle \int{x^2\cos{2x}\,dx} = \frac{1}{2}x^2\sin{2x} - \int{x\sin{2x}\,dx}$.

Now we need to try to find $\displaystyle \int{x\sin{2x}\,dx}$.

We should differentiate the other term in our original equation:

$\displaystyle \frac{d}{dx}(2x\cos{2x}) = 2\cos{2x} - 4x\sin{2x}$.

So $\displaystyle \int{2\cos{2x}\,dx} - \int{4x\sin{2x}\,dx} = 2x\cos{2x}$

$\displaystyle 4\int{x\sin{2x}\,dx} = 2\int{\cos{2x}\,dx} - 2x\cos{2x}$

$\displaystyle \int{x\sin{2x}\,dx} = \frac{1}{2}\int{\cos{2x}\,dx} - \frac{1}{2}x\cos{2x}$

$\displaystyle \int{x\sin{2x}\,dx} = \frac{1}{4}\sin{2x} - \frac{1}{2}x\cos{2x}$.

So... going back to what we were originally trying to find

$\displaystyle \int{x^2\cos{2x}\,dx} = \frac{1}{2}x^2\sin{2x} - \int{x\sin{2x}\,dx}$

$\displaystyle = \frac{1}{2}x^2\sin{2x} + \frac{1}{2}x\cos{2x} - \frac{1}{4}\sin{2x} + C$.

7. Originally Posted by Prove It
I should have read the question properly.

As an alternative to Integration by parts, there's the "Find the derivative and create a new integral equation" method.

$\displaystyle \frac{d}{dx}(x^2\cos{2x}) = 2x^2\sin{2x} + 2x\cos{2x}$.

So $\displaystyle \int{2x^2\sin{2x}\,dx} + \int{2x\cos{2x}\,dx} = x^2\cos{2x}$

Now to find these integrals, differentiate each of them...

$\displaystyle \frac{d}{dx}(2x^2\sin{2x}) = 4x^2\cos{2x} + 4x\sin{2x}$

So $\displaystyle \int{4x^2\cos{2x}\,dx} + \int{4x\sin{2x}\,dx} = 2x^2\sin{2x}$

Therefore $\displaystyle 4\int{x^2\cos{2x}\,dx} = 2x^2\sin{2x} - 4\int{x\sin{2x}\,dx}$

$\displaystyle \int{x^2\cos{2x}\,dx} = \frac{1}{2}x^2\sin{2x} - \int{x\sin{2x}\,dx}$.

Now we need to try to find $\displaystyle \int{x\sin{2x}\,dx}$.

We should differentiate the other term in our original equation:

$\displaystyle \frac{d}{dx}(2x\cos{2x}) = 2\cos{2x} - 4x\sin{2x}$.

So $\displaystyle \int{2\cos{2x}\,dx} - \int{4x\sin{2x}\,dx} = 2x\cos{2x}$

$\displaystyle 4\int{x\sin{2x}\,dx} = 2\int{\cos{2x}\,dx} - 2x\cos{2x}$

$\displaystyle \int{x\sin{2x}\,dx} = \frac{1}{2}\int{\cos{2x}\,dx} - \frac{1}{2}x\cos{2x}$

$\displaystyle \int{x\sin{2x}\,dx} = \frac{1}{4}\sin{2x} - \frac{1}{2}x\cos{2x}$.

So... going back to what we were originally trying to find

$\displaystyle \int{x^2\cos{2x}\,dx} = \frac{1}{2}x^2\sin{2x} - \int{x\sin{2x}\,dx}$

$\displaystyle = \frac{1}{2}x^2\sin{2x} + \frac{1}{2}x\cos{2x} - \frac{1}{4}\sin{2x} + C$.

Quite nice. good show.

8. Originally Posted by oblixps
this was a question on a calc test.

Let F(x) be the antiderivative of f(x) = (x^2) cos(2x) and F(1) = 0
a) find F(x) and F(3)
b) find the critical values of F(x) in [0,pi]
c) find F''(x)
d) For the values you found in b), explain which (if any) are local maxima. Justify your answer.
e) Find the absolute maximum value of F(x).

now these questions are pretty straightforward and easy but the problem i have is that there is no graphing calculator allowed on this test. i answered a,b,c, and, d easily, but e) is giving me some trouble. i found the local max at 3pi/4 and the two endpoints 0 and pi i have to test as well by plugging back into F(x). now this test was before we learned integration by parts so we can't use that technique. so the only way to find F(x) was to use the fundamental theorem of calculus so that F(x) = integral (from 1 to x) of (t^2)cos(2t) dt and evaluating F(x) means evaluating the respective integrals. the problem i have is that i don't know how to evaluate these integrals without a graphing calculator. My TI-84 has a program for evaluating definite integrals but the scientific calculators provided for us has no such function. how are you supposed to do part e) using only a scientific calculator?
I'm not sure that you need to check the points 0 and $\displaystyle \pi$ It sounds like from the question that that was only relevant for part b. If you're question is about part e) just set the equation you have equal to 0 and use F(x) from part a to decide what the absolute maximum is. (you should only have to try $\displaystyle x = 0$,$\displaystyle x = \pi/4$, and $\displaystyle x=3\pi/4$)