can anybody pls. help me with this math problem? im really stumped. thanks.?

Question

and can you tell me how to solve it so that i can understand it well.pls. click on the link for the math problem. thanks again.

problem no. 1. Evaluate the integral.
http://i12.photobucket.com/albums/a247/butteredbeer/582125img1.gif

Answer 1

♠ thus y(s) = -2exp(s)*sin(t-s); x= s-t, ds=dx; thus
y(x) = 2exp(x+t) *sin(x) =2exp(t) * exp(x)*sin(x); and limits for x=-t until 0;
I’d advise to use complex numbers knowing that exp(j*x) =cos(x) +j*sin(x);
since factor 2exp(t) is constant here consider:
z(x) = exp(x)*exp(j*x) = exp(x(1+j)); now integrate:
♣ Z(x) = exp(x*(1+j)) /(1+j) = exp(x)* exp(j*x) *(1-j)/2 =
= 0.5exp(x) *(cos(x) +j*sin(x)) *(1-j);
remembering we involved real part into z(x) now we have to get rid of it taking imaginary part into account:
♦ Y(x)= Im[Z(x)] = 0.5*exp(x) * (sin(x) –cos(x));
= {for x=-t until 0} = 0.5* {-1 -exp(-t) * (sin(-t) –cos(-t))};
recalling the factor we left in (♠) we receive:
♥ Y(t) = 2exp(t) *0.5 *{-1 +exp(t) *(sint +cost)} =
= sint +cost –exp(t); if I messed or for more details click me.

Answer 2

You integrate that by parts. However, there's a special trick to it. After you do it once, you will still be left with an integral. You do integration by parts on that again.

Now you will be left with an integral again. However, except for some multiplier, that integral is the same as the original integral. So what you have on the left of the equation is the original integral. On the right hand side of the equation you have the result of integration by parts performed twice. You now solve for the integral using algebra.

Once you solved for the integral, you evaluate the limits.

This is one of those special cases examples that they always teach you when they teach integration by parts.

Read this page http://en.wikipedia.org/wiki/Integration_by_parts

Somewhere in the middle of the page is an example very similar to yours where the expression to be integrated is an exponential times a trig function.

Answer 3

start up via dispensing the three into the x-4. To distribute, you're taking the quantity on the outdoors of the parentheses (3) and multiply it via the separate values on the interior (x) (-4). three times x equals 3x. three times -4 is -12. integrate the values. you're actually left with 3x-12. Your equation now sounds like this: x + 8 = 3x - 12 Now you may desire to get all the x values on an identical area: x + 8 = 3x - 12 -1x -1x 8 = 2x - 12 (From here i anticipate you be attentive to a thank you to sparkling up it, yet nonetheless) +12 + 12 20 = 2x /2 /2 10=x nicely, this is the respond. Sorry if I at a loss for words you.