Compute an area enclosed by the curve of

y=dxxln(x)y=dxxln⁡(x)

from x=2x=2 to x=4x=4


Question 5Answer

a.

ln(2)ln⁡(2)


b.

ln(16)ln⁡(16)


c.

ln(4)ln⁡(4)


d.

ln(1)ln⁡(1)


e.

ln(8)


The area under the curve \( y = x \ln(x) \) from \( x = 2 \) to \( x = 4 \) can be computed using the definite integral of \( y \) with respect to \( x \):

\[
A = \int_{2}^{4} x \ln(x) \, dx
\]

### Step 1: Apply integration by parts
To solve this, we will use integration by parts, where \( u = \ln(x) \) and \( dv = x \, dx \).

- Let \( u = \ln(x) \), so \( du = \frac{1}{x} dx \).
- Let \( dv = x \, dx \), so \( v = \frac{x^2}{2} \).

Now, apply the formula for integration by parts:

\[
\int u \, dv = uv - \int v \, du
\]

\[
\int x \ln(x) \, dx = \frac{x^2}{2} \ln(x) - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx
\]

This simplifies to:

\[
\frac{x^2}{2} \ln(x) - \int \frac{x}{2} \, dx
\]

### Step 2: Solve the integral
Now solve the remaining integral:

\[
\int \frac{x}{2} \, dx = \frac{x^2}{4}
\]

Thus, the result of the integration is:

\[
\frac{x^2}{2} \ln(x) - \frac{x^2}{4} + C
\]

### Step 3: Evaluate the definite integral
Now we evaluate the definite integral from \( x = 2 \) to \( x = 4 \):

\[
A = \left[ \frac{x^2}{2} \ln(x) - \frac{x^2}{4} \right]_{2}^{4}
\]

Substitute the upper and lower limits:

For \(

Compute the area enclosed by the curve y = x ln(x) from x = 2 to x = 4.

Learn how to compute the area under the curve y = x ln(x) from x = 2 to x = 4 using integration by parts. This problem explores calculus concepts like definite integrals and the fundamental theorem of calculus, making it suitable for students in Grades 11-12 or college level.

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