How do you evaluate the definite integral #int (e^(2x)) dx# from #[0,1]#?

Question

Samantha Adkison · Accepted Answer

Using the Fundamental Theorem of Calculus,
#int_0^1 e^(2x)dx=[e^(2x)/2]_0^1=1/2[e^2-e^0]#
#=1/2(e^2-1).#

Paisley Johnson · Answer

undefined To evaluate the definite integral $\int_0^1 e^{2x} \, dx$, you can use the following steps:

1. Find the antiderivative of $e^{2x}$ with respect to $x$, which is $\frac{1}{2} e^{2x}$.

2. Evaluate the antiderivative at the upper and lower limits of integration.

3. Subtract the value of the antiderivative at the lower limit from the value at the upper limit.

4. Simplify the result.

Applying these steps:

1. The antiderivative of $e^{2x}$ is $\frac{1}{2} e^{2x}$.

2. Evaluate the antiderivative at the upper limit:
   $\frac{1}{2} e^{2(1)} = \frac{1}{2} e^2$

Evaluate the antiderivative at the lower limit:
   $\frac{1}{2} e^{2(0)} = \frac{1}{2} e^0 = \frac{1}{2}$

3. Subtract the lower limit value from the upper limit value:
   $\frac{1}{2} e^2 - \frac{1}{2}$

4. Simplify the result if needed.

Therefore, the value of the definite integral $\int_0^1 e^{2x} \, dx$ is $\frac{1}{2} (e^2 - 1)$.

Christopher Agresta · Answer

undefined To evaluate the definite integral of $ \int_{0}^{1} e^{2x} \, dx $:

Integrate $ e^{2x} $ with respect to $ x $:
$$ \int e^{2x} \, dx = \frac{1}{2} e^{2x} + C $$

Evaluate the integral from 0 to 1:
$$ \left[ \frac{1}{2} e^{2x} \right]_{0}^{1} $$

Substitute the upper limit:
$$ \frac{1}{2} e^{2(1)} = \frac{1}{2} e^{2} $$

Substitute the lower limit:
$$ \frac{1}{2} e^{2(0)} = \frac{1}{2} e^{0} = \frac{1}{2} $$

Subtract the values at the lower limit from the upper limit:
$$ \frac{1}{2} e^{2} - \frac{1}{2} $$

$$ = \frac{1}{2} (e^{2} - 1) $$

So, $ \int_{0}^{1} e^{2x} \, dx = \frac{1}{2} (e^{2} - 1) $.