Question

find area of region y=xsin(x^2) on the interval [o,pi]

Answer 1

u know u should integrate for finding area right?

Answer 2

yeah but its been a minute since ive done this and im extremely rusty

Answer 3

i know you use f(a) - f(b)

Answer 4

emm..this is what we have\[\int_{0}^{\pi} x\sin x^2 \ \text{d}x\]

Answer 5

can you show me the steps? is it simply f(a)-f(b) of x - f(a)-f(b) of sinx^2?

Answer 6

I think u-sub works here, right?

Answer 7

He didn't say under the curve, so I'm assuming you'd need to break that integral up into 3 parts, and take the abs value of one part because it goes below x axis

Answer 8

well this is a u-sub integration let\[u=x^2\]\[du=2x\text{d}x\]\[\int_{0}^{\pi^2} \frac{1}{2} \sin u \text{d}u\]

Answer 9

ahh...thats a good point @Agent47 mentioned

Answer 10

@Agent47 always makes good points :D!

Answer 11

its just find the area enclosed by that equation. honestly i dont know where to begin to solve

Answer 12

but id say its just integrating because its impossible to solve \[x\sin x^2=0\]

Answer 13

exactly. I was going to use integration by parts and set uprime to \[\sin^{2}\]
and v = x , vprime=1

Answer 14

by parts should work too,

Answer 15

It's more difficult, though . .

Answer 16

u-substitution is short and sweet.

Answer 17

im not familiar with u-substitution

Answer 18

Check @mukushla 's third reply; he sets it up for you.

Answer 19

ohhhhhhhh

Answer 20

Isn't that a lot prettier than an integration-by-parts mess?

Answer 21

so completely bypass integration by parts and just go straight into solving the definite integral, subbing u out for pi and 0 and computing f(a) - f(b)???

Answer 22

yeah much much prettier

Answer 23

i still havent found the answer

Answer 24

its not \[\frac{ 1 }{ 2 }-\cos \int\limits_{0}^{\pi}\] is it?

Answer 25

im sorry cosu

Answer 26

Ok, start with the substitution like Mukushla showed:
Let u = x^2, so du = 2x dx

Answer 27

x dx = half of du, so your integral becomes
\[\int\limits x \sin(x^2) dx \rightarrow \frac{1}{2} \int\limits \sin(u) du\]

Answer 28

the integral of sin(u) is -cos(u), so you have
\[-\frac{1}{2} \cos (u), u=x^2 \rightarrow -\frac{1}{2} \cos (x^2)\]
Evaluate from 0 to π and you're done.

Answer 29

Be careful with the limits. I usually like to leave it indefinite until the end and then undo the substitution so I can keep the same limits, but really
\[\int\limits_{0}^{π} x \sin(x^2) dx = \frac{1}{2}\int\limits_{0}^{π^2}\sin(u) du, \space u=x^2\]

Answer 30

So the final answer should read...

\[-\frac{ 1 }{ 2 }\cos(\pi ^{2}) -(-\frac{ 1 }{ 2 }\cos(0) )\]

Answer 31

which gives -1/2 * 1 - (-1/2 * 1)???

Answer 32

which is 0.. which cant be right

Answer 33

Depends . . If you are subtracting areas below the x-axis, then maybe. If you add all the absolute values together, it should be something closer to 1.

Answer 34

But the way I solved it is right though?

Answer 35

Hmm, what's cos(π^2)?

Answer 36

cos(pi) is -1, which squared would equal 1. My calculator read -.9026853619

Answer 37

Square the π before taking cosine.

Answer 38

(order of operations . . .)

Answer 39

i still get the same answer

Answer 40

You mean -.9026853619, right?

Answer 41

yes

Answer 42

Ok, then go from there.

Answer 43

.951342681?

Answer 44

Yeah, that's what my calculator is telling me too.

Answer 45

Lol alright cool... thankyou

Answer 46

You're welcome.

I'm still curious if it was meant to subtract the areas below the x-axis or to add the absolute values of the areas..

I'm also confused why @mukushla said x*sin(x^2)=0 was impossible to solve; I found
x=0, x=sqrt(π), x=sqrt(2π), and x=sqrt(3π) on that interval.

Using those, you could set up four intervals of integration to find the positive areas of all the regions, . .
but perhaps that's more work than it's worth.

Answer 47

ahh...man my brain not functioning well these days .. u r right @CliffSedge .. its possible