Question

Sketch the region enclosed by y=e3x,y=e5x and x=1. Decide whether to integrate with respect to x or y, and then find the area of the region.

Answer 1

The area of the region is $\text { Area } \approx 23.12$

Explanation:

The area of the region is given by

$\text { Area }=\int_{0}^{1}\left(e^{5 x}-e^{3 x}\right) d x$

Integrating the terms inside the bracket, we get,

$\left[\frac{e^{5 x}}{5}-\frac{e^{3 x}}{3}\right]$

Now, applying the limits for x for the integrated term,

$Area=\frac{e^{5}}{5}-\frac{e^{3}}{3}-\frac{e^{0}}{5}+\frac{e^{0}}{3}$

Any number to the power of zero is one. Thus, substituting the values, we get,

$Area=\frac{e^{5}}{5}-\frac{e^{3}}{3}-\frac{1}{5}+\frac{1}{3}$

Adding the like terms,

$Area=\frac{e^{5}}{5}-\frac{e^{3}}{3}+\frac{2}{15}$

Simplifying the terms, we get,

$\text { Area } \approx 23.12$

Thus, The area of the region is $\text { Area } \approx 23.12$

Answer 2

The area of the region is 23.12 square unit

Given that:

$y = e^{3x}\\\\y = e^{5x}\\\\x = 1$

Find:

The area of the region

Computation:

$Area = \int\limits^1_0 {e^{5x} - e^{3x}} \, dx \\\\Area = \frac{e^5}{5} -\frac{e^3}{3} -[\frac{1}{5} - \frac{1}{3} ] \\\\Area = 29.68 - 6.69 +\frac{2}{15}\\\\Area = 22.99 + 0.133\\\\Area = 23.12$

The area of the region is 23.12 square unit

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