Find the area between the curve # y = e^(2x) # y = 0 from x = 1 to 3 ?

Question

Find the area between the curve # y = e^(2x) #  y = 0 from x = 1 to 3 ?

Samantha Adkison · Accepted Answer

Area #= 1/2 (e^6-e^2) = 198.019868 ... #

The shaded are sought is given by the integral:

# A = int_1^3 \ e^(2x) \ dx #
# \ \ \ = [ \ e^(2x)/2 \ ]_1^3 #
# \ \ \ = 1/2 (e^6-e^2) #
# \ \ \ = 198.019868 ... #

Christopher Agresta · Answer

undefined To find the area between the curve $ y = e^{2x} $ and the x-axis from $ x = 1 $ to $ x = 3 $, you need to compute the definite integral of $ e^{2x} $ with respect to $ x $ over the interval $[1, 3]$:

$$ \text{Area} = \int_{1}^{3} e^{2x} \, dx $$

To integrate $ e^{2x} $, use the following antiderivative:

$$ \int e^{ax} \, dx = \frac{1}{a} e^{ax} + C $$

where $ a $ is a constant. In this case, $ a = 2 $. So,

$$ \int e^{2x} \, dx = \frac{1}{2} e^{2x} + C $$

Evaluate the definite integral:

$$ \text{Area} = \left[\frac{1}{2} e^{2x}\right]_{1}^{3} $$
$$ = \left(\frac{1}{2} e^{6}\right) - \left(\frac{1}{2} e^{2}\right) $$
$$ = \frac{1}{2} (e^{6} - e^{2}) $$

Therefore, the area between the curve $ y = e^{2x} $ and the x-axis from $ x = 1 $ to $ x = 3 $ is $ \frac{1}{2} (e^{6} - e^{2}) $.

Luke Alvarez · Answer

undefined To find the area between the curve y = e^(2x) and the x-axis from x = 1 to x = 3, you need to integrate the absolute value of the function y = e^(2x) with respect to x over the interval [1, 3]. This is because the function e^(2x) is always positive for real values of x, so the area between the curve and the x-axis will be the integral of e^(2x) from x = 1 to x = 3.

∫[1 to 3] |e^(2x)| dx = ∫[1 to 3] e^(2x) dx

To integrate e^(2x), you can use the substitution method. Let u = 2x, then du = 2 dx.

∫ e^(2x) dx = (1/2) ∫ e^u du = (1/2) e^u + C

Now substitute back for u:

= (1/2) e^(2x) + C

Evaluate this expression from x = 1 to x = 3:

= (1/2) [e^(6) - e^(2)]

So, the area between the curve y = e^(2x) and the x-axis from x = 1 to x = 3 is (1/2) [e^(6) - e^(2)].