As part of a larger problem I need to find the definite integral from -4 to 0 of the function sqrt(16-x^2). I know that I'm overlooking something simple here because I've done harder problems without this much difficulty.

I've tried to solve it by substitution by letting u= 16-x^2

then du=-2xdx

-1/2du=xdx

 

and from there I don't know how to manipulate the function to substitute the -1/2du. I think that simplifying the function is probably the way to go, but I don't see how to go about it.

 

Thanks in advance.

My solution relies on a piece of knowledge I'm not sure you'll have, if you don't I've done something wrong.

 

Have you done complex numbers yet? (Are you expected to know about them?)

No, I would guess that my teacher would expect us to solve the problem by substitution or simplification (possibly by multiplying the function by a "carefully chosen value of 1", if that makes any sense to you).

Well I've just checked the answer (using an integral table), and I can't think of any simple substitution right now that would get it to work, sorry

Have you learned Trig. Substitutions?

Sorry, but based on what I can tell from the problem it looks like it is in the form of Trig substitutions. In which case, check them out. You'll see that the problem is in the format of sqrt(a^2-x^2). HINT: Let x= asin(theta)

I worked out part of the problem using that, and I believe it works out.

Let me know how it works out for you.

I haven't learned trig substitutions, but I gave it a go and only got so far...I realized that I didn't necessarily have to find out that integral in order to solve the larger problem at hand. Thanks for the help!

In case you're still interested in how it works...

 

[math]

\int_{-4}^{0} \sqrt{16 - x^2} \,dx

[/math]

 

We let [math] x= 4sin(\theta) [/math], therefore [math] dx= 4cos(\theta) d\theta [/math]

 

Now when we plug in these values we get [math] \int_{a}^{b} \sqrt{16-16sin(\theta)^2 }*4cos(\theta) d\theta [/math]. I put a and b for the endpoints because they will change twice throughout the entire problem and we can plug in the values when we get to the end.

 

Remembering our trig identities we now have [math] 16 \int_{a}^{b} \sqrt{1- sin(\theta)^2}* cos(\theta) d\theta = \int_{a}^{b} (cos(\theta))^2 d\theta [/math]

 

Next we use a half-angle formula to get [math] 8 \int_{a}^{b} 1+ cos(2\theta) d\theta [/math].

 

Now we can use [math] u-substitution [/math] letting [math] u= 2\theta [/math] and [math] du= 2 d\theta [/math] .

 

Finally we get [math] 4 \int_{a}^{b} 1+ cos(u) du = 4 [u+ sin(u)]^b_a= [ 2\theta+ sin(2\theta)]^b_a [/math]. We know [math] \theta= arcsin(x/4) [/math] so we plug in each original endpoint for x to get the new endpoints, and evaluate the function using the fundamental theorem of calculus. I got 12.6 for my answer.

Thanks, I appreciate it.