Question

evaluate the indefinite integral -2t + 2 (sin n pi t)dt

Answer 1

\(\Large \color{royalblue}{\text{Welcome to OpenStudy! :)}}\)

Answer 2

\[\Large \int\limits -2t+2 \sin\left(n \pi t\right)\;dt\]So this is what we need to solve?

Answer 3

can you help me how to solve if the question is definite integrals to the interval 1 until 1/2?

Answer 4

\[\Large \int\limits\limits_{1/2}^1 -2t+2 \sin\left(n \pi t\right)\;dt\]Like this? :o

Answer 5

\[\int\limits_{1}^{1/2} -2x + 2 (\sin n \pi t) dt\]
yess, of course

Answer 6

I'm sorry, like this \[\int\limits_{1/2}^{1} -2t + 2 (\sin n \pi t) dt\]

Answer 7

So for the first term, we simply apply the `Power Rule for Integration`, yes?\[\Large -2t \qquad\to\qquad \frac{-2}{2}t^2\]That part make sense? :o

Answer 8

Can you explain this problem with partial integral method?

Answer 9

partial integral method..? :o

Answer 10

hmmm, integral by part

Answer 11

Hmm there is no parts needed here :(

Answer 12

are you sure? how about \[\int\limits_{-1/2}^{1/2} 2t (\sin n \pi t) dt\] and my friend find the result is \[- \frac{ 2 \cos 1/2 n \pi }{ n \pi} + \frac{ 4 \sin 1/2 n \pi }{ n \pi ^{2} }\]. he used \[u v - \int\limits v du\]. what do you think?

Answer 13

\[\int\limits\limits_{-1/2}^{1/2} 2t (\sin n \pi t) dt\]This? Yes we can do parts on this :)