the region enclosed by
y
=
e
2
x
,
y
=
e
4
x
, and
x
=
1
. integrate with respect to
y
. Then find the area of the region.

To solve the problem of finding the area of the region enclosed by the curves \( y = e^{2x} \), \( y = e^{4x} \), and \( x = 1 \) using integration with respect to \( y \), follow these steps:

### 1. **Express \( x \) in terms of \( y \)**:
   For the given functions:
   \[
   y = e^{2x} \implies x = \frac{1}{2} \ln(y)
   \]
   \[
   y = e^{4x} \implies x = \frac{1}{4} \ln(y)
   \]

### 2. **Determine the bounds for \( y \)**:
   When \( x = 1 \):
   - For \( y = e^{2x} \), \( y = e^{2(1)} = e^2 \)
   - For \( y = e^{4x} \), \( y = e^{4(1)} = e^4 \)
   
   Therefore, the region is bounded by \( y = e^2 \) and \( y = e^4 \).

### 3. **Set up the integral**:
   The area \( A \) can be found by integrating the difference between the \( x \)-values of the curves (i.e., \( x = \frac{1}{2} \ln(y) \) and \( x = \frac{1}{4} \ln(y) \)) with respect to \( y \) from \( y = e^2 \) to \( y = e^4 \).

   The integral is:
   \[
   A = \int_{e^2}^{e^4} \left[\frac{1}{2} \ln(y) - \frac{1}{4} \ln(y)\right] dy
   \]
   Simplify the integrand:
   \[
   A = \int_{e^2}^{e^4} \left(\frac{1}{4} \ln(y)\right) dy
   \]

### 4. **Compute the integral**:
   The integral \( \int \ln(y) \, dy \) can be computed using integration by parts:
   \[
   \text{Let } u = \ln(y) \text{ and } dv = dy
   \]
   Then, \( du = \frac{1}{y} dy \) and \( v = y \).

   The integral becomes:
   \[
   \int \ln(y) \, dy = y \ln(y) - \int y \cdot \frac{1}{y} dy = y \ln(y) - y
   \]

   Therefore:
   \[
   A = \frac{1}{4} \left[ y \ln(y) - y \right]_{e^2}^{e^4}
   \]

### 5. **Evaluate the definite integral**:
   Substitute the bounds \( y = e^2 \) and \( y = e^4 \) into the expression:
   \[
   A = \frac{1}{4} \left[ (e^4 \ln(e^4) - e^4) - (e^2 \ln(e^2) - e^2) \right]
   \]

   Simplifying further using \( \ln(e^a) = a \):
   \[
   A = \frac{1}{4} \left[ (4e^4 - e^4) - (2e^2 - e^2) \right]
   \]
   \[
   A = \frac{1}{4} \left[ 3e^4 - e^2 \right]
   \]
   \[
   A = \frac{3}{4}e^4 - \frac{1}{4}e^2
   \]

So, the area of the region is \( \boxed{\frac{3}{4}e^4 - \frac{1}{4}e^2} \).

Would you like further details or have any questions?

### Related Questions:
1. How would you integrate with respect to \( x \) instead of \( y \) for the same region?
2. Can you find the centroid of the enclosed region?
3. How would you compute the volume if this region were revolved around the x-axis?
4. What would be the effect on the area if the region were enclosed by \( x = 2 \) instead of \( x = 1 \)?
5. How does the choice of integrating with respect to \( y \) instead of \( x \) simplify or complicate the problem?

### Tip:
Always consider the symmetry and bounds of the region before deciding the variable of integration. This can simplify the computation process significantly.

The region enclosed by y = e^(2x), y = e^(4x), and x = 1. Integrate with respect to y. Then find the area of the region.

Learn how to calculate the area of the region enclosed by the curves y = e^(2x), y = e^(4x), and x = 1 using integration with respect to y. This step-by-step guide includes key calculus concepts such as logarithmic functions and integration by parts. Suitable for advanced high school or college-level students.

Math Problem Statement

the region enclosed by y

e 2 x , y

e 4 x , and x

Solution

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