Question

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

Answer 1

`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:

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`

Answer 2

`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`

Answer 3

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.