Answer to Question #180264 in Calculus for Prathamesh

Question

Evaluate the area between y = x2 and the line y = 2x.

Answer 1

To find the area under the curve $y=x^2 \text{ and } y=2x$ , we sketch the curve.

to get the upper limit and lower limit $[a,b]$ of the functions, we get the point of intersection.

Since $y=x^2 \text{ and } y=2x$ , then

$$x^2=2x\\ \implies x^2-2x=0\\ x(x-2)=0\\ x=0 \text{ or } x-2=0\\ \therefore x= 0,2$$

We proceed to integrate the function about the limits so as to get the area under the curve:

$$A= \int_{0}^{2} (2x-x^2)\,dx\\ = \Bigg [x^2-\frac{x^3}{3} \Bigg ]^2_1\\ \qquad \qquad \quad= \Bigg [2^2-\frac{2^3}{3} \Bigg ]-\Bigg [0^2-\frac{0^3}{3} \Bigg ]\\ = 4-\frac{8}{3}\\ A = 2.67 \text{ sq.units}$$