Definite Integral of a Trig Function

[Jason76]

\(\displaystyle \int^{0}_{1}\sin x\)

Any hints? I know how to do definite integrals, that is, without trig functions. I do know the integral of \(\displaystyle \sin (x)\) is \(\displaystyle -\cos(x)\). Do I do the "integral power rule" on this one?

 

[DrPhil]

\(\displaystyle \int^{0}_{1}\sin x\)

Any hints? I know how to do definite integrals, that is, without trig functions. I do know the integral of \(\displaystyle \sin (x)\) is \(\displaystyle -\cos(x)\). Do I do the "integral power rule" on this one?

NO

Please forget you ever heard the words, "integer power rule."

If you know the integral, use it.

 

[Jason76]

I see what's going on know. I think in that situation there was too much thinking. Basically, doing integrals is about "taking the integral". Take the integral of the trig function and you have the answer. Next, evaluate at the upper and lower bounds. Now, there are situations using the integral power rule, but not in this case.

Ok here is the answer:

\(\displaystyle \int^{0}_{1}\sin(x) = -\cos(x) + C\)

\(\displaystyle -\cos(0) - (-\cos(1)) = -\cos(0) + \cos(1) = -1 + -.5403 = -1.5483\)

 

[Deleted member 4993]

I see what's going on know. I think in that situation there was too much thinking. Basically, doing integrals is about "taking the integral". Take the integral of the trig function and you have the answer. Next, evaluate at the upper and lower bounds. Now, there are situations using the integral power rule, but not in this case.

Ok here is the answer:

\(\displaystyle \int^{0}_{1}\sin(x) = -\cos(x) + C\)

\(\displaystyle -\cos(0) - (-\cos(1)) = -\cos(0) + \cos(1) = -1 + -.5403 = -1.5483\) .... Incorrect

It should be:

\(\displaystyle -\cos(0) - (-\cos(1)) = -\cos(0) + \cos(1) = -1 + 0.5403 = -0.4597\)

.