Question

The area bounded by \(y=x^2\) and \(y=2x\) from \(x=0\) to \(x=2\) is:

• A. \(\frac{8}{3}\)
• B. \(\frac{4}{3}\)
• C. \(\frac{16}{3}\)
• D. \(\frac{10}{3}\)

Hint:

Upper curve minus lower curve gives area.

Answer 1

The Correct Option is A

To find the area bounded by the curves \(y = x^2\) and \(y = 2x\) from \(x = 0\) to \(x = 2\), we need to calculate the area between these two curves over the interval.

First, determine the points of intersection by setting the equations equal to each other:

\(x^2 = 2x\)

Rearrange this to:

\(x^2 - 2x = 0\)

Factor the equation:

\(x(x - 2) = 0\)

Thus, the points of intersection are \(x = 0\) and \(x = 2\).

Next, calculate the area between the two curves using definite integrals:

The area \(A\) between two curves \(y_1(x)\) and \(y_2(x)\) from \(x = a\) to \(x = b\) is given by:

\(A = \int_a^b [y_2(x) - y_1(x)] \, dx\)

In this problem, from \(x = 0\) to \(x = 2\), the curve \(y = 2x\) is above \(y = x^2\). Thus:

\(A = \int_0^2 (2x - x^2) \, dx\)

Compute the integral:

\(A = \left[ x^2 - \frac{x^3}{3} \right]_0^2 = \left(2^2 - \frac{2^3}{3}\right) - \left(0 - \frac{0}{3}\right)\)

Simplify further:

\(A = \left(4 - \frac{8}{3}\right) = \frac{12}{3} - \frac{8}{3} = \frac{4}{3}\)

Thus, the area between the curves from \(x = 0\) to \(x = 2\) is \(\frac{8}{3}\).

Correct Answer: \(\frac{8}{3}\)