Question

Given Integral of sint/(t^2+t+1)dt from 0 to 3/x, Calculate G'(x).

Answer 1

\[G(x)=\int\limits_{0}^{3/x}\sin t/ (t^2+t+1) dt\]

Answer 2

now when we differentiate G(x) you first plug in the upper limit in the function and then differentiate the limit minus the lower limit plugged in and it's differentiation.
since lower limit is 0 its differentiation will result in 0 so only upper limit is needed

Answer 3

check out the procedure
\[G'(x)=d/dx \int\limits_{0}^{3/x}\sin t/(t^2+t+1) dt\]
\[G'(x)=\sin(3/x)/((3/x)^2+3/x+1) d/dx(3/x)-0\]
solution
\[G'(x)=(-3/x^2)\sin(3/x)/((3/x)^2+3/x+1) \]