Question

Sketch the region bounded by the graphs of the functions, and find the area of the region. $$ f(x)=x e^{-x^{2}}, \quad y=0, \quad 0 \leq x \leq 1 $$

Short answer

The area under the curve \(f(x) = x e^{-x^2}\) from \(x=0\) to \(x=1\) is \(\frac{2-e}{2e}\).

Step-by-step solution

Sketch the given function.

To gain a more comprehensive understanding of the question, sketch the given function \(f(x) = x e^{-x^2}\) for the provided interval, which is \(0 \leq x \leq 1\). In this region the function is positive and the graph lies above the x-axis.

Identify the bounded region.

The bounded region is the area between the curve \(f(x) = x e^{-x^2}\) and the x-axis (y=0) from \(x=0\) to \(x=1\). This is the area we are asked to compute.

Compute the area under the curve.

To find the area under a curve from \(x=a\) to \(x=b\), we can use the definite integral from \(a\) to \(b\) of the function. Apply this concept to the bounded region from Step 2, the integral to compute the area under the curve is \(\int_{0}^{1} x e^{-x^{2}} dx\). This can be found using a simple substitution \(u=x^2\), \(du=2x dx\) and changing the bounds accordingly. The resulting integral \(\int_{0}^{1} \frac{e^{-u}}{2} du\) can be integrated using standard techniques of calculating the integral of exponential functions.

Evaluate the definite integral.

After performing the calculations from the previous steps, we get \(\left. -\frac{e^{-u}}{2} \right|_0^1 = -\frac{1}{2e} + \frac{1}{2} = \frac{2-e}{2e}\).