Compute the following integral: $\int _{\ln3}^{\ln 6}\:8e^xdx$

Question

Compute the following integral:

$$\int _{\ln3}^{\ln 6}\:8e^xdx$$

My attempt

We have $$\begin{align}\int \:8e^xdx=8\cdot \int \:e^xdx=8e^x+C\end{align}$$

Now $$\lim _{x\to \ln (3)^+}\left(8e^x\right)=8e^{\ln(3)}=8\cdot3=24$$

and $$\lim _{x\to \ln (6)^-}\left(8e^x\right)=8e^{\ln(6)}=8\cdot6=48$$

Hence $$\int _{\ln3}^{\ln 6}\:8e^xdx=48-24=24$$

How did I do?

Answer 1

You can just use the Fundamental Theorem of Calculus: $$\int_a^b f(x)\,dx=F(b)-F(a)$$ where $F(x)$ is the antiderivative of $f(x)$. You also shouldn't use a constant when evaluating a definite integral.

In your case, \begin{align} 8\int_{\ln 3}^{\ln 6}e^x\,dx&=8\cdot e^x\Bigg|_{\ln 3}^{\ln 6} \\ &=8\cdot (6-3) \\ &=24 \end{align}

Answer 2

\begin{align} \int_{\ln 3}^{\ln 6} 8e^x \, dx &= [8e^x]_{\ln 3}^{\ln 6} \\ &=8(\exp(\ln6)-\exp(\ln3)) \\ &=8(6-3)=8(3)=24 \end{align}

I notice that you write equality at every single line such as "$=6$" on one line follow by "$=6 \cdot 8$", it can carry misguiding information.