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How do you integrate #int 3 dt#?

1 Answer

Feb 26, 2017

#3t+C.#

Explanation:

We know that, for a const. #k, intkf(t)dt=kintf(t)dt.#

#:. int3dt=3intdt=3intt^0dt.#

Since, #intt^ndt=t^(n+1)/(n+1)+c, where, n!=-1,# we have,

#int3dt=3intt^0dt=3{t^(0+1)/(0+1)}=3(t^1/1)=3t+C.#

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