Question

Difficult integral

Answer 1

\[\int\limits_{3}^{-3} (x+3)\sqrt{9-x^{2}} dx\]

Answer 2

@poopsiedoodle

Answer 3

@inkyvoyd

Answer 4

Try some u sub.

Answer 5

I'd be more specific, but you should let us know what you're struggling about.

Answer 6

maybe fundamental theorem of calculus? because.. the f(b) should be greater than the f(a)

Answer 7

put x= 3 sin t

Answer 8

you could simplify the integral, by using
\(\int \limits_{a}^b f(x)dx = \int \limits_a^b f(a+b-x)dx \)

then you'll only have
\(\int \limits_{-3}^3 3 \sqrt{9-x^2}dx\)
which is not as difficult!

Answer 9

\(I= \int\limits_{3}^{-3} (x+3)\sqrt{9-x^{2}} dx ...(A) \\ I= \int\limits_{3}^{-3} (-x+3)\sqrt{9-x^{2}} dx ...(B) \\ Add \: (A) , (B) \\...\)

Answer 10

this is also equivalent to removing odd part of the function for calculating the intergal with limits -a, to a.
because \(\int \limits_{-a}^a odd\: function \:dx= 0\)

Answer 11

@hartnn Can we solve this integral using integration by parts

Answer 12

i think yes, but it will be more complicated..

Answer 13

My first impulse would be to re-write the integral as follows:\[\int\limits\limits_{3}^{-3} (x+3)\sqrt{9-x^{2}} dx=-\int\limits\limits_{-3}^{3} (x+3)\sqrt{9-x^{2}} dx\]

Answer 14

Next, I'd separate this single integral into 2 separate integrals, in which x would multiply the square root in the first such integral and 3 would multiply the square root in the second.

Trig substitution would likely work, too. What such substitution would be appropriate here when we have 9 - x^2 under the radical sign?

Answer 15

\[\int\limits_{3}^{-3} (x+3)\sqrt{9-x^{2}} dx = \int_{-3}^3 x \sqrt{9-x^2} dx + \int_{-3}^3 \sqrt{9-x^2} dx\]

Geometrically the first integral's function is odd so the section on (-3,0) is the negative of the area on (0,3) so they cancel each other out. This just leaves us with this integral:

\[ \int_{-3}^3 \sqrt{9-x^2} dx\]

if you look, \(y=\sqrt{9-x^2}\) is really \(x^2+y^2 = 3^2\). So this function is a circle, and we're finding the area of a circle that's above the x-axis, so it's just:

\[\frac{\pi 3^2}{2}\]

Answer 16

Hey, K! That's pretty perceptive. But having alternative ways to solve one problem can be a good thing.

Answer 17

Yeah I'm not saying my way is the only way! My favorite part about math is collecting multiple ways of thinking and I like questions that people have already answered in alternate ways as a challenge to see if I can come up with some other way of doing it that no one else has.

Answer 18

K, you missed the 3, but nice way! :)

Answer 19

Thanks guys
@Kainui
We can solve the second integral using trig substitution. I was stuck with the integral until I realized trig substitution works. Thanks though