How do you evaluate the integral int x/(sqrt(16-x^2)dx from 0 to 4?

3 Answers
Jul 24, 2016

int_(0)^(4) (x)/(sqrt(16-x^2)) dx = 4

Explanation:

For this particular integral, we can use a trigonometric substitution.

If we have the form sqrt(a^2-u^2) we can do the following:

u=asintheta

du = acostheta d theta

sqrt(a^2-u^2) = acostheta

If it helps, you can use a triangle to to visualize this:

enter image source here

Since we have the integral int_(0)^(4) (x)/(sqrt(16-x^2)) dx

Let a = 4, so u = 4sintheta -> du = 4costheta d theta

Also note that u/4 = sintheta

Therefore, also changing the limits of integration, we can write

x = 4sintheta

x = 0 -> theta = 0

x = 4 -> theta = pi/2

int_(0)^(4) (x)/(sqrt(16-x^2)) dx = int_(0)^(pi/2) (4sintheta)/(4costheta) 4costheta d theta = int_(0)^(pi/2) 4sintheta d theta

Remembering that int sin theta d theta = -costheta + C, we get

int_(0)^(pi/2) 4sintheta d theta = -4[costheta]_(0)^(pi/2)

By looking at the triangle, we can see that costheta = sqrt(16-x^2)/(4), so

-4[costheta]_(0)^(pi/2) = -4*[(sqrt(16-x^2))/(4)]_(0)^(4)

= -4[sqrt(16-16)/(4) - sqrt(16-0)/(4)]

= -4[0-4/4] = -4[-1] = 4

Jul 25, 2016

4

Explanation:

Although it is a possibility, a trigonometric substitution is not necessary.

This can also be tackled using the substitution u=16-x^2. This implies that du=-2xdx. Thus:

int_0^4x/sqrt(16-x^2)dx=-1/2int_0^4(-2x)/sqrt(16-x^2)dx

Before making the u and du substitutions, recall that the bounds will change. Plug the current bounds into 16-x^2. Thus the bound of 0 becomes 16-0^2=16 and the bound of 4 becomes 16-4^2=0.

-1/2int_0^4(-2x)/sqrt(16-x^2)dx=-1/2int_16^0 1/sqrtudu

From here, we can reorder the integral using the rule: int_a^bf(x)dx=-int_b^af(x)dx. Also, rewrite 1/sqrtu using fractional and negative exponents:

-1/2int_16^0 1/sqrtudu=1/2int_0^16u^(-1/2)du

From here, integrate using the rule: intu^ndu=u^(n+1)/(n+1) and then evaluate the integral.

1/2int_0^16u^(-1/2)du=1/2[u^(-1/2+1)/(-1/2+1)]_0^16=1/2[u^(1/2)/(1/2)]_0^16

The 1/2 can be brought from the denominator as a 2, cancelling with the 1/2 lingering outside of the brackets, leaving:

1/2[u^(1/2)/(1/2)]_0^16=[sqrtu]_0^16=sqrt16-sqrt0=4

Jul 25, 2016

as a mere third approach, but one where you can pretty much do the heavy lifting in your head, provided you're up for a bit of pattern recognition....

we have int_0^4 x/(sqrt(16-x^2))dx

and from the power rule d/dx x^n = n x^(n-1), with a bit of chain rule thrown in too, we can see that

d/dx( sqrt(16-x^2) )= 1/2 1/sqrt(16-x^2) (- 2x) = -x/sqrt(16-x^2)

IOW!!
- d/dx( sqrt(16-x^2) )= x/sqrt(16-x^2) which is our integrand

so int_0^4 x/(sqrt(16-x^2))dx = int_0^4 - d/dx ( sqrt(16-x^2) ) \ dx

= - [ sqrt(16-x^2) ]_0^4 = [ sqrt(16-x^2) ]_4^0 = 4

whilst the other 2 answers have been beautifully presented and are very elegant, my suggestion is this .... well, they didn't say you had to use a trig sub here, or any sub for that matter, and you don't have to so why bother?

and these patterns show up all over the place.