Question

Write an integral in the form $A = \int_a^b h(x) \, dx$ such that $A$ expresses the area of the region between the functions $y = e^x$, $y = e^{2x-1}$ and $x = 0$. Evaluate the integral to find the area of the region. Answers should be exact. As a suggestion, graph the equations so you can see the region for which you are solving for the area. a (lower limit of integration) = 0 b (upper limit of integration) = $\frac{1}{3}$ $h(x) = e^{2x} - e^{2x - 1}$ Area of region $A = \frac{3}{10}e^{\frac{2}{3}} - \frac{1}{2} + \frac{1}{10}$

          Write an integral in the form $A = \int_a^b h(x) \, dx$ such that $A$ expresses the area of the region between the functions $y = e^x$, $y = e^{2x-1}$ and $x = 0$. Evaluate the integral to find the area of the region. Answers should be exact. As a suggestion, graph the equations so you can see the region for which you are solving for the area.
a (lower limit of integration) = 0
b (upper limit of integration) = $\frac{1}{3}$
$h(x) = e^{2x} - e^{2x - 1}$
Area of region $A = \frac{3}{10}e^{\frac{2}{3}} - \frac{1}{2} + \frac{1}{10}$

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Write an integral in the form A = ^b h(x) dx such that A expresses the area of the region between the functions y = e^x, y = e^2x-1 and x = 0. Evaluate the integral to find the area of the region. Answers should be exact. As a suggestion, graph the equations so you can see the region for which you are solving for the area.
a (lower limit of integration) = 0
b (upper limit of integration) = (1)/(3)
h(x) = e^2x - e^2x - 1
Area of region A = (3)/(10)e^(2)/(3) - (1)/(2) + (1)/(10)

Added by Danny A.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Vincenzo Zaccaro

08/06/2022

Step 1

Since the region is between the functions y = -1 and y = 0, we need to find the x-values where these functions intersect. Setting y = -1 and y = 0 for each function, we get: -1 = -1 and 0 = -1 So, the x-values where the functions intersect are at y = -1. Show more…

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When an integral in the form A = ∫h dx such that A expresses the area of the region between the functions y = -1 and y = 0. Evaluate the integral to find the area of the region. Answers should be exact. As a suggestion, graph the equation so you can see the region for which you are solving for the area. H 2r_5t1 Area of region A

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