Question

NEED CALCULUS HELP!

so I can solve c and d on a practice test using an integral and my calculator but I cannot figure out HOW to solve the actual integral by hand- which is what my teacher grades on.

c) region, R, is bound by y=x^3 - 4x and y= sin(pi x). R is the base of a solid with square cross sections perpendicular to the x-axis. What is that figures volume?

d) given the same R as being the surface of a pond and the knowledge that the ponds depth always = 3-x, meaning it gets more shallow as x grows and is the same depths for all values of y with a similar x. What is the pond's volume?

Answer 1

Welcome to OpenStudy!
I think it will be so much better if you had post your answer choices this way. (Example)
A) 234
B) 123
C) 10,000
D) 1

Answer 2

I don't have answer choices. He just told us to answer the problem. Did you maybe just mean the questions?

Answer 3

Never mind then.......

Answer 4

Did they give bounds for problem c) ?

Answer 5

sorry, 0 to 2

Answer 6

i know it is INT ( (sin (pi x) - x^3 +4x)^2) from 0 to 2 but I don't know how to solve it by hand

Answer 7

I tried factoring sin (pi x) - x^3 +4x)^2 and then splitting the answer into multiple integrals so that I could do substitution but it gave me a different answer than my calculator- which is correct according to the AP site

Answer 8

I would expand out the quadratic

Answer 9

do you mean factoring (sin (pi x) - x^3 +4x)^2) so it is sin^2(pi x) - x^3 sin(pi x) + 4x sin(pi x) - x^3 sin(pi x) - x^6 -8x^4 + 4x sin(pi x) + 16x^2 ?

Answer 10

Yes, but I don't call that factoring.

we can integrate the powers of x terms easily enough

the 8x sin(pi x) we can integrate by parts
sin^2(pi x) we can use the identity
\[ sin^2(\pi x)= \frac{1}{2} (1- \cos(2 \pi x) ) \]

Answer 11

the hardest term is -2x^3 sin(pi x)

Answer 12

for sin^2 (pi x) i had been doing int( sin^2(pi x) = - (1/pi) cos(pi x)
was that wrong?

Answer 13

yes. if we think of it as
\[ u^2 du \]
with u = sin(x), then du = cos(x) dx
we don't have a cos(x) dx so we can't do that.
we use a trig substitution

Answer 14

I am sorry, but where did the pi*x go?

Answer 15

oh, I was being sloppy. Use
\[ \sin^2(\pi x)= \frac{1}{2} (1- \cos(2 \pi x) ) \]

Answer 16

why is that true?

Answer 17

cos(a+b)= cos(a)cos(b) - sin(a)sin(b)
if a=b, we get cos(2a) = cos^2(a) - sin^2(a)
replace cos^2(a) with 1 - sin^2(a)
cos(2a) = 1 - sin^2(a)- sin^2(a)
do some algebra

Answer 18

thanks

Answer 19

we can integrate
x^3 sin(pi x) using integration by parts a few times in a row. Painful, but doable.

Answer 20

i finally got c! thanks!

Answer 21

mathnerd99 wrote:

I'm looking back at what you've said trying to figure this out and so I have 2 questions:

1) when you said earlier sin2(πx)=1/2(1−cos(2πx)), this was in reference to 8x*sin(pi*x), this is an actual identity? also just making sure- nothing there was integrated or derived, right?

The 8x sin(pi x) is not relevant here. To integrate \( \sin^2(\pi x) \) we use the identity to change it to a form we can integrate.



2)why do is an equation for cos(a+b) necessary? is it just to get to the bit about cos(2a) = 1 - sin^2(a)- sin^2(a)? would a= pi*x ?

Yes. The "double angle" formula for cos(2a) follows from the sum of angles formula, and I started as far back as seemed reasonable... if you had also wanted a derivation for the sum of angles cos(a+b), I would have given you a link, because it takes too long to write out.

After some work, I found the volume to be
\[ 1 -\frac{24}{\pi^3} +\frac{128}{7} - \frac{256}{5} + \frac{128}{3} \approx 9.978\]