Question

integrate 36e^x sinx dx from -infinity to 0

Answer 1

you mean:
\[\int\limits_{-\infty}^{0}36e^xsinx dx\]

Answer 2

Yes. I understand how to find the ∫e^xsinx dx. I understand the integration by parts. I understand the algebra required to solve ∫e^xsinx dx. I am making a mistake somewhere in my integration..
\[\lim_{a \rightarrow -\infty} 72\int\limits_{a}^{0} e ^{x}sinx dx = \lim_{a \rightarrow -\infty}36e ^{x}(sinx-cosx) from a \to 0\]

Answer 3

The integral of\[\int\limits_{}^{}e^{x}\sin(x) = \frac{ -\cos(x)e^{x}+\sin(x)e^{x} }{ 2 }\]
Did you get this result?

Replacing that into the original you should get:

\[36( \frac{ -\cos(x)e^{x}+\sin(x)e^{x} }{ 2 }) = \frac{ -\cos(x)e^{x}+\sin(x)e^{x} }{ 18 }\]

Answer 4

if i remove limit, then...

Integration by parts, twice:

u = sin x; dv = e^x dx
du = cos x dx; v = e^x

36[∫ e^x sin x dx] = 36[e^x sin x - ∫ e^x cos x dx]

u = cos x; dv = e^x dx
du = - sin x dx; v = e^x

36[∫ e^x sinx dx] = 36[e^x sin x - { e^x cos x - ∫ e^x (- sin x) dx}]

36[∫ e^x sinx dx] = 36[e^x sin x - e^x cos x - ∫ e^x sin x dx]

Bring the original integral to the same side by adding and factor the right side:
72[ ∫ e^x sin x dx] =36[ e^x ( sin x - cos x)]

Divide by 2:
36[∫ e^x sin x dx] =18[ e^x (sin x - cos x)]

Answer 5

I made a huge mistake at the end of my post (shouldm't have divided by 18). @bittuaryan has explained the process well.

Answer 6

Forgive me I am new to this. I apologize for having 15 years since my last algebra class. I see where my division choice was incorrect.
Evaluating from -infinity to 0 I get
=\[\lim_{a \rightarrow -\infty} 18e^{0}(\sin0-\cos0)-18e^{a}(\sin(a)-\cos(a)) \]
=\[18(1)(0-1)-18e^{-\infty}(\sin(-\infty)-\cos(-\infty)) \] How does this go to 18 ?

Answer 7

Thank you @DeoxNa and @bittuaryan for explaining everything thus far. Does anyone know how the limit goes to -18 from -infinity to 0.

How does \[18e^{−∞}(\sin(−∞)−\cos(−∞)) = 0\]

Answer 8

\[18e ^{-\infty}(\sin(-\infty)-\cos(-\infty))=0\]
Because

\[e ^{-\infty}=0\]
As x approaches infinity, e^x gets negligibly close to zero.

Answer 9

Thank you.

Answer 10

You're welcome :)

Answer 11

not mention, dear @LACHEEK989