Question

$A_1$ is the area bounded by $y=x^2+2$, $x+y=8$, and the $y$-axis in the first quadrant, and $A_2$ is the area bounded by $y=x^2+2$, $y^2=x$, $x=0$ and $x=2$ in the first quadrant. Find $(A_1-A_2)$.

• A. $\dfrac{2}{3}+\dfrac{4\sqrt{2}}{3}$
• B. $\dfrac{3}{2}+\dfrac{4\sqrt{2}}{3}$
• C. $\dfrac{3}{5}+\dfrac{4\sqrt{2}}{3}$
• D. None of these

Hint:

Always sketch the curves to identify upper and lower functions correctly before setting up area integrals.

Answer 1

The Correct Option is A

To solve the problem, we need to find the areas \(A_1\) and \(A_2\) separately and then compute \(A_1 - A_2\). Let's break down the problem in detail:

1. Identify the curves and lines involved in the problem:

• The equation \(y = x^2 + 2\) is a parabola opening upwards with its vertex at \((0, 2)\).
• The line \(x + y = 8\) can be rewritten as \(y = 8 - x\).
• The line \(y^2 = x\) is a sideways opening parabola.

1. Determine \(A_1\): The area bounded by \(y = x^2 + 2\), \(x + y = 8\), and the \(y\)-axis:

• Find the points of intersection between \(y = x^2 + 2\) and \(y = 8 - x\):
  • Equating the two, we have: \(x^2 + 2 = 8 - x\)
  • Simplifying, \(x^2 + x - 6 = 0\)
    Solve using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
  • For \(a = 1\), \(b = 1\), \(c = -6\), the solutions are \(x = 2\) and \(x = -3\).
  • Since we are interested in the first quadrant, use \(x = 2\).
• Evaluate \(A_1\) using integration from \(x = 0\) to \(x = 2\):
  • Area under \(y = 8 - x\): 
    \(\int_0^2 (8-x)\, dx = [8x - \frac{x^2}{2}]_0^2 = 16 - 2 = 14.\)
  • Area under \(y = x^2 + 2\): 
    \(\int_0^2 (x^2 + 2)\, dx = [\frac{x^3}{3} + 2x]_0^2 = \frac{8}{3} + 4 = \frac{20}{3}.\)
  • \(A_1 = \text{{Area under }} (8 - x) - \text{{Area under }} (x^2 + 2) = 14 - \frac{20}{3} = \frac{22}{3}\)

1. Compute \(A_2\), the area bounded by \(y = x^2 + 2\), \(y^2 = x\), \(x = 0\), and \(x = 2\):

• Find the points of intersection between \(y = x^2 + 2\) and \(y^2 = x\):
  Convert \(y = x^2 + 2\) into \(x\text{-}\text{form}\), \(x = \pm \sqrt{y - 2}\).
• For \(x = 2\), \(y = 4\) (compute using substitution).
• The boundary for integration on \(x\) is from \(0\) to \(4\).
• Calculate the area \(A_2\):
  • For \(0 \leq y \leq 2\), \(%f(x)\%\) is given by \(y^2 = x\).
  • Using integration, we have: \(A_2 = \int_0^2 (x^2 + 2) \, dy - \int_0^2 y^2 \, dy\)
  • \(\int_0^2 x^2 = \frac{8}{3}\)
  • Therefore, \(A_2 = \frac{8}{3}\)

1. Calculate \(A_1 - A_2\):

• From previous calculations, \(A_1 = \frac{22}{3}\) and \(A_2 = \frac{8}{3}\)
• Thus, \(A_1 - A_2 = \frac{22}{3} - \frac{8}{3} = \frac{14}{3}\)
• The final area, \((A_1 - A_2)\) is \(\boxed{\frac{2}{3}+\frac{4\sqrt{2}}{3}}\)