Evaluating Def. Integrals: int[1,2][(x-1)sqrt(2-x)]dx, etc

[2000WS6]

I believe my main issue here is integrating the equations I have here. I've gone to the tutoring center at my college to no avail (idiot just started talking to some chick for over half an hour so I walked) - he did however mention that the only way he knew how to solve equations like this is by Integration by Parts - which we have yet to learn in Calc 1. Please help me solve these. We've been mainly using u-substitution to solve our integrals.

These are towards the end of the section and I haven't really been having much trouble solving these but I can't even pull \(\displaystyle u\) and \(\displaystyle du\) out of these equations to save my life. With this one, it's the \(\displaystyle (x - 1)\) that's throwing me off.


\(\displaystyle \int_1^2 (x - 1) \sqrt{2 - x} dx\)


One more I can't wrap my head around - I've worked it out more than once but the answer is WAY off.

\(\displaystyle \int_0^\frac{\pi}{2} cos(\frac{2x}{3}) dx\)


Also, I have a relatively small favor to ask. I'm not quite sure the concept they're trying to convey here but I can't seem to figure out a logical way to solve these problems using the "Symmetry of the graphs of sine and consine functions as an aid in evaluating each of the integrals" (as quoted from the directions). Any aid would be appreicated

\(\displaystyle a) \int_\frac{-\pi}{4}^\frac{\pi}{4} sin(x) dx\) Now, I know this equals 0 because of sine's behavior between these two points, creating equal areas under the x-axis and over the x-axis canceling out to zero. The rest I have trouble with because I'm unsure how to determine area by way of the directions.

\(\displaystyle b) \int_\frac{-\pi}{4}^\frac{\pi}{4} cos(x) dx\)

I won't list the rest, as I'm confident that if I can see how that one's done, I'll be able to do the rest. Thank you very much for your help and I'm sorry if I've asked anything painfully obvious. The end of the semester is fast approaching and I tend to freak out and psyche myself out at times. Your time is appreciated.

- Ryan

 

[soroban]

Re: Evaluating the Definate Integral

Hello, 2000WS6!

\(\displaystyle \int_1^2 (x - 1) \sqrt{2 - x}\, dx\)

\(\displaystyle \text{Let: }\,u \:=\:\sqrt{2-x}\quad\Rightarrow\quad u^2\:=\:2-x\quad\Rightarrow\quad x \:=\:2-u^2\)

. . \(\displaystyle x - 1 \:=\:1-u^2\quad\Rightarrow\quad dx \:=\:-2u\,du\)

\(\displaystyle \text{Substitute: }\;\int(1-u^2)\cdot u (-2u\,du) \;\;=\;\;-2\int(u^2-u^4)\,du\)

. . . . . \(\displaystyle =\;-2\left(\frac{u^3}{3} - \frac{u^5}{5}\right) \;\;=\;\;-\frac{2}{15}u^3(5-3u^2)\)

\(\displaystyle \text{Back-substitute: }\;-\frac{2}{15}(2-x)^{\frac{3}{2}}\bigg[5-3(2-x)\bigg] \;\;=\;\; -\frac{2}{15}(2-x)^{\frac{3}{2}}(3x-1)\:\bigg]^2_1\)

\(\displaystyle \text{Evaluate: }\;-\frac{2}{15}(2-2)^{\frac{3}{2}}(6-1) + \frac{2}{15}(2-1)^{\frac{3}{2}}(3-1) \;\;=\;\;0 + \frac{2}{15}(1)(2) \;=\;\boxed{\frac{4}{15}}\)

 

[galactus]

One more I can't wrap my head around - I've worked it out more than once but the answer is WAY off.

\(\displaystyle \int_0^\frac{\pi}{2} cos(\frac{2x}{3}) dx\)


Let \(\displaystyle u=\frac{2}{3}x, \;\ \frac{3}{2}du=dx\)


\(\displaystyle \frac{3}{2}\int{cos(u)}du\)

\(\displaystyle =\frac{3}{2}sin(u)\)

Resub:

\(\displaystyle \frac{3}{2}sin(\frac{2x}{3})\)

Now, use the limits of integration and you're done.

 

[skeeter]

since cosine is an even function ...

\(\displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos{x} \, dx = 2\int_{0}^{\frac{\pi}{4}} \cos{x} \, dx\)

 

[2000WS6]

Thank you all very much for your help.

1st one was cooky as far as what I've seen in my current calc class. That's a lot of figuring with \(\displaystyle u\) to get that one to work out.

2nd one is PAINFULLY obvious. I'm sorry I didn't see that and posted it up - thx for taking the time to point out my ignorant mistake.

3rd one was also painfully obvious after my teacher finished up the lecture today. No biggie there.

Thank you all very much for your time. I do appreciate it!