Home > Calculus > Evaluating the Constant of Integration

What is the integral of 5(6t+e^(-t))(2t+4) with respect to t?

#int(5(6t+e^-t)(2t+4))dt#

Answer 1

Isaiah Allen

#5int(6t+e^-t)(2t+4)dt=20t^3+60t^2-10te^-t-30e^-t+C#

Factor out #5# and multiply through.

#5int(6t+e^-t)(2t+4)dt#

#5int(12t^2+24t+2te^-t+4e^-t)dt#

We can split this sum up and integrate each term individually.

#5(int12t^2dt+int24tdt+int2te^-tdt+int4e^-tdt)#

#int12t^2dt=12/3t^3=4t^3#

#int24tdt=24/2t^2=12t^2#

#int4e^-tdt=-4e^-t#

So, the more complicated integral is #int2te^-tdt.# We'll have to integrate by parts.

#u=2t#

#du=2dt#

#dv=e^-tdt#

#v=-e^-t#

#uv-intvdu=-2te^-t+2inte^-tdt=-2te^-t-2e^-t#

Combining all of the above:

#5int(6t+e^-t)(2t+4)dt=5(4t^3+12t^2-2te^-t-2e^-t-4e^-t)+C#

Don't forget the constant of integration. Simplifying yields

#5int(6t+e^-t)(2t+4)dt=20t^3+60t^2-10te^-t-30e^-t+C#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

William Abdalla

The integral of (5(6t+e^{-t})(2t+4)) with respect to (t) is:

[ \int 5(6t+e^{-t})(2t+4) , dt = \frac{15}{2}t^2 + 20t + 10e^{-t} + C ]

where (C) is the constant of integration.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.