Integration by Parts: int_0^4 [ e^(-x) sin(5pi x) ] dx

[canaanbowman]

$$\int_0 ^4[e^{-x} \sin(5\pi x)]dx$$
I integrated by parts and got this for an indefinite integral:

$$F(x)=-\frac{1}{26}[5e^{-x}\cos(5\pi x)+e^{-x}\sin(5\pi x)]+C$$
When I evaluate the definite integral in the original problem by finding $$F(4)-F(0)$$
I get 0.188785.


However, when I evaluate the integral directly using NINT on a graphing calculator I get 0.0622437.

I have double checked my integral by differentiating and it comes out right. Am I missing something or could the integral be giving the calculator a stroke?

 

[nasi112]

Did you check the angle set to radians instead of degrees in the calculator?

My guess is telling me that you did something wrong while integrating by parts.

 

[canaanbowman]

Did you check the angle set to radians instead of degrees in the calculator?

My guess is telling me that you did something wrong while integrating by parts.

I did check that it was in radians. I can find no error in the integrating by parts. When you take the derivative of what I have for F(X) it goes right back to the expression in the integrand.

 

[blamocur]

Try differentiating your $F(x)$ to see if you got it right. Hint: you did not.
As for the numerical value, it looks correct to me.

 

[blamocur]

When you take the derivative of what I have for F(X) it goes right back to the expression in the integrand.

No, it does not. Hint: $\frac{d}{dx} \sin 5\pi x \neq 5 \cos 5 \pi x$.

 

[canaanbowman]

I did check that it was in radians. I can find no error in the integrating by parts. When you take the derivative of what I have for F(X) it goes right back to the expression in the integrand.

Did you check the angle set to radians instead of degrees in the calculator?

My guess is telling me that you did something wrong while integrating by parts.

Here is my work for integrating by parts using the formula: $$\int u\ dv=uv-\int v\ du$$
$$\int e^{-x}\sin({5\pi x})dx$$
$$u=e^{-x}\ \ \ \ \ du=-e^{-x}\ dx\ \ \ \ \ dv=\sin(5\pi x)\ dx\ \ \ \ \ v=-\frac{1}{5}\cos(5\pi x)$$
$$\int e^{-x}\sin({5\pi x})dx=-\frac{1}{5}e^{-x}\cos(5\pi x)-\int\frac{1}{5}\cos(5\pi x)e^{-x}dx$$
Again using the formula for the integral at the end:

$$u=\frac{1}{5}e^{-x}\ \ \ \ \ du=-\frac{1}{5}e^{-x}dx\ \ \ \ \ dv=\cos(5\pi x)\ dx\ \ \ \ \ v=\frac{1}{5}\sin(5\pi x)$$
$$\int e^{-x}\sin(5\pi x)=-\frac{1}{5}e^{-x}\cos(5\pi x)-\bigg[\frac{1}{25}e^{-x}\sin(5\pi x)-\int -\frac{1}{25}e^{-x}\sin(5\pi x)dx]\bigg]$$
$$\int e^{-x}\sin(5\pi x)=-\frac{1}{5}e^{-x}\cos(5\pi x)-\frac{1}{25}e^{-x}\sin(5\pi x)-\frac{1}{25}\int e^{-x}\sin(5\pi x)dx$$
$$\frac{26}{25}\int e^{-x}\sin(5\pi x)=-\frac{1}{5}e^{-x}\cos(5\pi x)-\frac{1}{25}e^{-x}\sin(5\pi x)$$
$$\int e^{-x}\sin(5\pi x)=-\frac{5}{26}e^{-x}\cos(5\pi x)-\frac{1}{26}e^{-x}\sin(5\pi x)+C$$
$$\int e^{-x}\sin(5\pi x)=-\frac{1}{26}\bigg[5e^{-x}\cos(5\pi x)+e^{-x}\sin(5\pi x)\bigg]+C$$

 

[canaanbowman]

No, it does not. Hint: $\frac{d}{dx} \sin 5\pi x \neq 5 \cos 5 \pi x$.

I see it now!! The PI!

Let this be a good and long lesson that pi is part of the constant!

Thank you so much, friend!