Question

Evaluate $\int^1_0 {e^{2-3x}} \, dx$ as a limit of a sum.

Answer 1

check the attachment

Answer 2

Answer:

$\dfrac {e^3 - 1}{3e}$

Step-by-step explanation:

We have to find

$\int^1_0 {e^{2-3x}} \, dx$

Let us find the solution

$=\int^1_0 e^2 \cdot e^{-3x} dx\\e^2 \int^1_0 e^{-3x} dx\\\\=e^2 \left [\dfrac {e^{-3x}}{-3}\right ]_0^1 \\\\=\dfrac {-e^2}3\left [ e^{-3\times 1}-e^{-3 \times 0} \right ]\\\\=\dfrac {-e^2}3\left [e^{-3} - 1\right ]\\\=\ \dfrac{e^2}3\left (1 - e^{-3}}\right )$

Now,

$=\dfrac {e^3 - 1}{3e}$

Therefore the answer is

$\int^1_0 {e^{2-3x}} \, dx=\dfrac {e^3 - 1}{3e}$