How do you calculate the value of the integral #inte^(4t²-t) dt# from #[3,x]#?

2 Answers
Aug 1, 2015

Answer:

#inte^(4t^2-t)dt=(e^(4x^2-x))/(8x-1)-e^(33)/23#

Explanation:

Be #f(x)=e^(4t^2-t)# your function.

In order to integrate this function, you will need its primitive #F(x)#

#F(x)=(e^(4t^2-t))/(8t-1)+k# with #k# a constant.

The integration of #e^(4t^2-t)# on [3;x] is calculated as follows:

#inte^(4t^2-t)dt=F(x)-F(3)#

#=(e^(4x^2-x))/(8x-1)+k-((e^(4cdot3^2-3))/(8cdot3-1)+k)#

#=(e^(4x^2-x))/(8x-1)-e^(33)/23#

Aug 1, 2015

Answer:

That integral cannot be expressed using elementary functions. If requires the use of #int e^(x^2) dx#. However the derivative of the integral is #e^(4x^2-x)#

Explanation:

The fundamental theorem pf calculus part 1 tells us that the derivative with respect to #x# of:

#g(x) = int_a^x f(t) dt# is #f(x)#

So the derivative (with respect to #x#) of

#g(x) = int_3^x e^(4t^2-t) dt" "# is #" "g'(x) = e^(4x^2 -x)#.