Evaluate the definite integral of 1.62482e^(-x/12.72238) + 2.85422 with respect to x from x = 0 to x = 300.

Step-by-step solution

1. 1

   Split the integral into the two pieces you can handle separately. The integrand is a decaying exponential plus a constant, and integration is linear, so

   $\int_0^{300}\!\!\left(a\,e^{-x/k}+c\right)dx=a\int_0^{300}\!e^{-x/k}dx+c\int_0^{300}\!dx$

   with $a=1.62482$, $k=12.72238$, $c=2.85422$.

2. 2

   Antidifferentiate the exponential. Since $\dfrac{d}{dx}e^{-x/k}=-\dfrac{1}{k}e^{-x/k}$, the antiderivative picks up a factor of $-k$:

   $\int a\,e^{-x/k}dx=-ak\,e^{-x/k}+C,\qquad \int c\,dx=cx+C$

   so $F(x)=-ak\,e^{-x/k}+cx$.

3. 3

   Evaluate at the limits and simplify.

   $F(300)-F(0)=ak\left(1-e^{-300/k}\right)+300c$

   The $-e^{-x/k}$ at the lower limit becomes $+1$, which is why the decay term contributes a positive $ak$.

4. 4

   Compute the numbers carefully. The decay coefficient is

   $ak=1.62482\times12.72238=20.6715774716$

   and the constant part is

   $300c=300\times2.85422=856.266$

5. 5

   Show the exponential tail is negligible. $\dfrac{300}{12.72238}=23.5804$, and $e^{-23.5804}=5.75\times10^{-11}$. So $1-e^{-300/k}=0.99999999994$, which changes $ak$ only in the eleventh decimal place — the integral has effectively reached its infinite-horizon value $ak+300c$.

6. 6

   Add and state the result, correcting a common slip.

   $20.6715774716+856.266=876.9375774705$

   A value of $876.93843$ is sometimes quoted, but it comes from mis-multiplying $1.62482\times12.72238$ as $20.6724315116$; the true product is $20.6715774716$.

7. 7

   Verify by numerical integration. A trapezoidal sum with three million subintervals over $[0,300]$ gives $876.9375775$, matching the closed form to seven decimals ✓.