Question

Graph the region between the curves and use your calculator to compute the area correct to five decimal places.

\(y = \frac{2}{{1 + {x^4}}},\quad y = {x^2}\)

Short answer

The area of the region is \(2.80120\).

Step-by-step solution

The area \(A\) of the region bounded by the curves is given by

The area \(A\) of the region bounded by the curves \(y = f(x),y = g(x)\), and the lines \(x = a,x = b\), where \(f\) and \(g\) are continuous and \(f(x) \ge g(x)\) for all \(x\) in (a, b), is \(A = \int_a^b {(f(x) - g(x))dx} \)

Sketch the graph of the curves

Sketch the graph of the curve \(y = \frac{2}{{1 + {x^4}}}\)and \(y = {x^2}\)as shown in the figure below

Procedure to sketch the region bounded by the two curves is explained below:

- Substitute different values for\(x\)and drawthe graph for the function\(y = \frac{2}{{1 + {x^4}}}\) .

- Substitute different values for\(x\)and plot for the function\(y = {x^2}\)

- Shade the region where the points intersect on the curves.

The region enclosed by the curves \(y = \frac{2}{{1 + {x^4}}}\) and \(y = {x^2}\) is shown in below.

Calculate the area of the region enclosed.

Evaluate the area bounded as follows

The curves intersect at\(x = - 1\)and\(x = 1\).

The curves are bounded by the top and bottom curve, so the integration can be done with respect to\(x\).

Find the area of the region bounded by the curves:

\(A = \int_a^b {(f(} x) - g(x))dx(1)\)

Here, the top curve is\(f(x)\), the bottom curve is\(g(x)\), the lower limit is\(a\), and the upper limit is\(b\).

Substitute\(\frac{2}{{1 + {x^4}}}\)for\(f(x),{x^2}\)for\(g(x), - 1\)for\(a\), and 1 for\(b\)in Equation (1).

\(A = \int_{ - 1}^1 {\left( {\frac{2}{{1 + {x^4}}} - {x^2}} \right)} dx{\rm{ }}\)

Solve the equation by the use of calculator.\(A = 2.80120\)

Therefore, the area of the region enclosed by the curves is\(2.80120\).