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.