Definte Integral for consumption of gas in ten years

[cmnalo]

How much gas will be used in a 10 year period
time(years)=t=0 corresponds to 2000

Total cosumtion of gas:

∫ (t + e^0.01t)dt interval [0,10]

u=0.01t

du=0.01dt

dt= du/0.01

∫ (t + e^u)dt

I'm confused on what to do next.

Answer: 50+100(e^0.1 -1) =approximately 60.5

 

[galactus]

Hey cmnalo:

This is a cut and dry integral.

\(\displaystyle \L\\\int_{0}^{10}(t+e^{\frac{t}{100}})dt\)

\(\displaystyle =\L\\\int_{0}^{10}tdt+\int_{0}^{10}e^{\frac{t}{100}}dt\)

Now, it's like falling off a "log". No pun intended. Well, maybe there was, corny as it may be.

 

[Unco]

You should recognise that \(\displaystyle \int e^{at} \, dt \, = \, \frac{1}{a}e^{at}\) (plus a constant).

If you wish to work through a substitution, first recall that you can integrate term by term:
\(\displaystyle \int t + e^{0.01t} \, dt = \int t \, dt \, + \, \int e^{0.01t} \, dt\)

Then, for the latter term, you have correctly u = 0.01t and dt = 1/0.01 du, so that \(\displaystyle \int e^{0.01t} \, dt = \frac{1}{0.01} \int e^{u} \, du\), after having brought out the constant factor \(\displaystyle \frac{1}{0.01}\).