Question

The area bounded by the curve $(y - x)^2 = x^3$ and the line $x = 1$ is A (in square units). Then the value of 10A is

Answer 1

8

Answer 2

The problem asks us to find the area bounded by the curve $(y - x)^2 = x^3$ and the line $x = 1$, and then calculate the value of $10A$, where $A$ is this area.

1. Express the curve in terms of y: The given equation is $(y - x)^2 = x^3$. Taking the square root on both sides, we get: $y - x = \pm \sqrt{x^3}$ $y - x = \pm x^{3/2}$ So, $y = x \pm x^{3/2}$.

This gives us two branches of the curve:

• $y_1 = x + x^{3/2}$ (upper branch)
• $y_2 = x - x^{3/2}$ (lower branch)

2. Determine the domain for x: For $\sqrt{x^3}$ to be a real number, $x^3$ must be non-negative. This implies $x \ge 0$.

3. Set up the integral for the area: The area $A$ is bounded by the curve and the line $x=1$. Since $x \ge 0$, the region of interest spans from $x=0$ to $x=1$. The area between two curves $y_1(x)$ and $y_2(x)$ from $x=a$ to $x=b$ is given by $\int_a^b |y_1(x) - y_2(x)| dx$. In our case, for $x \ge 0$, $x^{3/2} \ge 0$, so $y_1 = x + x^{3/2}$ is greater than or equal to $y_2 = x - x^{3/2}$. The difference between the two branches is: $y_1 - y_2 = (x + x^{3/2}) - (x - x^{3/2})$ $y_1 - y_2 = x + x^{3/2} - x + x^{3/2}$ $y_1 - y_2 = 2x^{3/2}$

The area $A$ is the integral of this difference from $x=0$ to $x=1$: $A = \int_0^1 (y_1 - y_2) dx = \int_0^1 2x^{3/2} dx$

4. Evaluate the integral: $A = 2 \int_0^1 x^{3/2} dx$ Using the power rule for integration, $\int x^n dx = \frac{x^{n+1}}{n+1}$: $A = 2 \left[ \frac{x^{3/2 + 1}}{3/2 + 1} \right]_0^1$ $A = 2 \left[ \frac{x^{5/2}}{5/2} \right]_0^1$ $A = 2 \left[ \frac{2}{5} x^{5/2} \right]_0^1$ $A = \frac{4}{5} [x^{5/2}]_0^1$ Now, apply the limits of integration: $A = \frac{4}{5} (1^{5/2} - 0^{5/2})$ $A = \frac{4}{5} (1 - 0)$ $A = \frac{4}{5}$ square units.

5. Calculate the value of 10A: The problem asks for the value of $10A$. $10A = 10 \times \frac{4}{5}$ $10A = 2 \times 4$ $10A = 8$

The final answer is $\boxed{8}$.