How do you find the integral #int_0^(2)(10x)/sqrt(3-x^2)dx# ?

Question

How do  you find the integral #int_0^(2)(10x)/sqrt(3-x^2)dx# ?

Paisley Johnson · Accepted Answer

This is a classic case if what's called u-substitution. Meaning that you have to find a function ( u) and it's derivative (du) in the expression. Both the function and it's derivative may be hidden behind coefficients and confusing notation.

In this case, I might start by using exponents to rewrite the integral without fractions.

#int_0^2(10x)/sqrt(3-x^2)dx=int_0^2(3-x^2)^(-1/2)(10x)dx#

Now, I'm looking for a function and it's derivative. Here's where I notice that I have a second degree polynomial ( #3-x^2#) inside the exponent part, and I have a first degree polynomial (#10x#) elsewhere in the integral. In general a polynomial that starts with an x^2 will have a derivative that starts with an x. So:

If I decide that

#u=3-x^2#

then

#du=-2xdx#

but I don't have a #-2x# anywhere on the integral. I do have a #10x#.

I'm allowed to manipulate coefficients in an integral (manipulating variables is trickier, and sometimes not possible). So I need to manipulate the #10# in front of the x into being a #-2# in front of an x.

We start with:

#int_0^2(3-x^2)^(-1/2)(10x)dx#

We can pull a coefficient factor out of the integral entirely. We don't have to, but I find it makes for a more clear picture.

#10int_0^2(3-x^2)^(-1/2)(x)dx#

Now we would really like to put a #-2# in front of the x, so that we can match the #-2x# above. And we can put in in there. As long as we put it's reciprocal out in front to balance things out.

#10(1/(-2))int_0^2(3-x^2)^(-1/2)(-2x)dx#

=#-5int_0^2(3-x^2)^(-1/2)(-2x)dx#

Now we have our u and our du. But there's one other thing we need to be aware of.

This is a definite integral, and the 0 and the 2 represent x values, not u values. But we have a way to convert them,

#u=3-x^2#

So for #x=0#, we get #u=3-0^2#, or #u=3#.

For #x=2#, we get #u=3-2^2#, or #u=-1#.

Now we substitute all our x's for u's.

#-5int_0^2(3-x^2)^(-1/2)(-2x)dx=-5int_3^(-1)u^(-1/2)du#

#=-5(u^(1/2))/(1/2)|_3^(-1)=-10u^(1/2)|_3^(-1)#

#=-10sqrt(-1)-(-10sqrt(3))#

#=-10i+10sqrt(3)#, or

#10(sqrt(3)-i)#

Hope this helps.

Axel Alvarez · Answer

undefined To find the integral $ \int_{0}^{2} \frac{10x}{\sqrt{3-x^2}} \, dx$, you can use the substitution method:

Let $ u = 3 - x^2 $.

Then, $ du = -2x \, dx $ and $ x \, dx = -\frac{1}{2} \, du $.

When $ x = 0 $, $ u = 3 - 0^2 = 3 $.

When $ x = 2 $, $ u = 3 - 2^2 = -1 $.

Now, we rewrite the integral with respect to $ u $:

$ \int_{3}^{-1} \frac{-10}{2} \frac{1}{\sqrt{u}} \, du $.

$ = -5 \int_{3}^{-1} u^{-\frac{1}{2}} \, du $.

$ = -5 \left[ \frac{u^{\frac{1}{2}}}{\frac{1}{2}} \right]_{3}^{-1} $.

$ = -10 \left[ \sqrt{u} \right]_{3}^{-1} $.

$ = -10 \left( \sqrt{-1} - \sqrt{3} \right) $.

$ = -10 \left( i\sqrt{1} - \sqrt{3} \right) $.

$ = -10i + 10\sqrt{3} $.

Therefore, the integral $ \int_{0}^{2} \frac{10x}{\sqrt{3-x^2}} \, dx = -10i + 10\sqrt{3} $.