Question

The area bounded by the curve y = x(2-x) and the line y=x is

• A. \(\frac{1}{6}\)
• B. \(\frac{1}{3}\)
• C. \(\frac{1}{2}\)
• D. \(\frac{5}{6}\)
• E. \(\frac{2}{3}\)

Answer 1

The Correct Option is A

We are given the curve \( y = x(2 - x) \) and the line \( y = x \). To find the area bounded by the curve and the line, we first need to find the points of intersection of these two functions.
Step 1: Find the points of intersection Set \( x(2 - x) = x \) to find the points of intersection: \[ x(2 - x) = x \] \[ 2x - x^2 = x \] \[ 2x - x^2 - x = 0 \] \[ -x^2 + x = 0 \] \[ x(x - 1) = 0 \] Thus, \( x = 0 \) and \( x = 1 \) are the points of intersection.
Step 2: Set up the integral for the area The area between the curve and the line from \( x = 0 \) to \( x = 1 \) is given by the integral: \[ \text{Area} = \int_0^1 \left[ x(2 - x) - x \right] dx \] Simplify the integrand: \[ x(2 - x) - x = 2x - x^2 - x = x - x^2 \] Thus, the area is: \[ \text{Area} = \int_0^1 (x - x^2) dx \]
Step 3: Compute the integral Integrate \( x - x^2 \): \[ \int (x - x^2) dx = \frac{x^2}{2} - \frac{x^3}{3} \] Now evaluate the definite integral: \[ \text{Area} = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - \left( \frac{0^2}{2} - \frac{0^3}{3} \right) \] \[ \text{Area} = \left( \frac{1}{2} - \frac{1}{3} \right) \] \[ \text{Area} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \]

The correct option is (A) : \(\frac{1}{6}\)

Answer 2

We want to find the area bounded by the curve \(y = x(2 - x)\) and the line \(y = x\).

First, we find the points of intersection by setting the two equations equal to each other: \[x(2 - x) = x\] \[2x - x^2 = x\] \[x^2 - x = 0\] \[x(x - 1) = 0\] So, \(x = 0\) or \(x = 1\). Thus, the points of intersection are (0, 0) and (1, 1).

Now, we set up the integral. We need to determine which function is greater on the interval [0, 1]. Let's test a point between 0 and 1, say \(x = \frac{1}{2}\): For \(y = x(2 - x)\), \(y = \frac{1}{2}\left(2 - \frac{1}{2}\right) = \frac{1}{2}\left(\frac{3}{2}\right) = \frac{3}{4}\) For \(y = x\), \(y = \frac{1}{2}\) Since \(\frac{3}{4} > \frac{1}{2}\), \(x(2 - x) > x\) on the interval (0, 1).

So, the area between the curves is: \[A = \int_{0}^{1} [x(2 - x) - x] \, dx = \int_{0}^{1} (2x - x^2 - x) \, dx = \int_{0}^{1} (x - x^2) \, dx\]

Now, we evaluate the integral: \[A = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_{0}^{1} = \left(\frac{1^2}{2} - \frac{1^3}{3}\right) - \left(\frac{0^2}{2} - \frac{0^3}{3}\right) = \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6}\]

Therefore, the area bounded by the curve \(y = x(2 - x)\) and the line \(y = x\) is \(\frac{1}{6}\).