Calculating Area $A_1$
Area $A_1$ is enclosed by $y = x^2 + 2$, $x + y = 8$ (or $y = 8 - x$), and the $y$-axis ($x=0$) in the first quadrant.
- Find intersection points: Set $x^2 + 2 = 8 - x$. This gives $x^2 + x - 6 = 0$, or $(x+3)(x-2) = 0$. In the first quadrant, the intersection is at $x=2$.
- Set up the integral: The upper curve is $y = 8 - x$ and the lower curve is $y = x^2 + 2$ between $x=0$ and $x=2$. $A_1 = \int_{0}^{2} [(8 - x) - (x^2 + 2)] dx$
- Calculate the integral: $A_1 = \int_{0}^{2} (6 - x - x^2) dx = \left[ 6x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{2}$ $A_1 = \left( 6(2) - \frac{2^2}{2} - \frac{2^3}{3} \right) - (0) = 12 - 2 - \frac{8}{3} = 10 - \frac{8}{3} = \frac{30 - 8}{3} = \frac{22}{3}$
Calculating Area $A_2$
Area $A_2$ is enclosed by $y = x^2 + 2$, $y^2 = x$ (or $y = \sqrt{x}$ in the first quadrant), $x = 2$, and the $y$-axis ($x=0$) in the first quadrant.
- Determine the upper and lower curves: For $x \in [0, 2]$, $x^2+2 \ge \sqrt{x}$. Thus, $y = x^2 + 2$ is the upper curve and $y = \sqrt{x}$ is the lower curve.
- Set up the integral: The area is between $x=0$ and $x=2$. $A_2 = \int_{0}^{2} [(x^2 + 2) - \sqrt{x}] dx$
- Calculate the integral: $A_2 = \int_{0}^{2} (x^2 + 2 - x^{1/2}) dx = \left[ \frac{x^3}{3} + 2x - \frac{x^{3/2}}{3/2} \right]_{0}^{2}$ $A_2 = \left[ \frac{x^3}{3} + 2x - \frac{2}{3}x^{3/2} \right]_{0}^{2}$ $A_2 = \left( \frac{2^3}{3} + 2(2) - \frac{2}{3}(2^{3/2}) \right) - (0)$ $A_2 = \frac{8}{3} + 4 - \frac{2}{3}(2\sqrt{2}) = \frac{8}{3} + \frac{12}{3} - \frac{4\sqrt{2}}{3} = \frac{20 - 4\sqrt{2}}{3}$
Calculating $A_1 - A_2$
Subtract the calculated areas $A_1$ and $A_2$. $A_1 - A_2 = \frac{22}{3} - \frac{20 - 4\sqrt{2}}{3}$
- Perform the subtraction: $A_1 - A_2 = \frac{22 - (20 - 4\sqrt{2})}{3} = \frac{22 - 20 + 4\sqrt{2}}{3} = \frac{2 + 4\sqrt{2}}{3}$
- Factor the result: $A_1 - A_2 = \frac{2(1 + 2\sqrt{2})}{3} = \frac{2}{3}(1 + 2\sqrt{2})$
This result matches option C, which is $\frac{2}{3}(2\sqrt{2} + 1)$.
Similar Questions
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- If $\int (\sin x)^{\frac{-11}{2}} (\cos x)^{\frac{-5}{2}} dx = - \frac{p_1}{q_1} (\cot x)^{\frac{9}{2}} - \frac{p_2}{q_2} (\cot x)^{\frac{5}{2}} - \frac{p_3}{q_3} (\cot x)^{\frac{1}{2}} + \frac{p_4}{q_4} (\cot x)^{\frac{-3}{2}} + C$, where $p_i$ and $q_i$ are positive integers with $\gcd(p_i, q_i) = 1$ for $i = 1, 2, 3, 4$ and C is the constant of integration, then $\frac{15 p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4}$ is equal to _________
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If $y = y(x)$ satisfies the differential equation
$16(\sqrt{x+ 9\sqrt{x}})(4 + \sqrt{9 + \sqrt{x}}) \cos y \, dy = (1 + 2 \sin y) dx, x > 0$ and $y(256) = \frac{\pi}{2}, y(49) = \alpha$, then $2 \sin \alpha$ is equal to :- Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f\left(\frac{tx}{36}\right) dt = 4\alpha f(x)$. If $y = f(x)$ is a standard parabola passing through the points (2, 1) and (– 4, $\beta$), then $\beta^\alpha$ is equal to ______.
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Let $f(x) = \int \frac{7x^{10} + 9x^8}{(1 + x^2 + 2x^9)^2} \,dx$, $x > 0$, $\lim_{x \rightarrow 0} f(x) = 0$ and $f(1) = \frac{1}{4}$.
If $A = \begin{bmatrix} 0 & 0 & 1 \\ \frac{1}{4} & f'(1) & 1 \\ \alpha^2 & 4 & 1 \end{bmatrix}$ and $B = \text{adj}(\text{adj } A)$ be such that $|B| = 81$, then $\alpha^2$ is equal to
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Important Questions from Integral Calculus
- Let the solution curve of the differential equation $x dy - y dx = \sqrt{x^2 + y^2} dx, x > 0$, $y(1) = 0$, be $y = y(x)$. Then $y(3)$ is equal to
- If $\int (\sin x)^{\frac{-11}{2}} (\cos x)^{\frac{-5}{2}} dx = - \frac{p_1}{q_1} (\cot x)^{\frac{9}{2}} - \frac{p_2}{q_2} (\cot x)^{\frac{5}{2}} - \frac{p_3}{q_3} (\cot x)^{\frac{1}{2}} + \frac{p_4}{q_4} (\cot x)^{\frac{-3}{2}} + C$, where $p_i$ and $q_i$ are positive integers with $\gcd(p_i, q_i) = 1$ for $i = 1, 2, 3, 4$ and C is the constant of integration, then $\frac{15 p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4}$ is equal to _________
- The area of the region $A = \{(x, y) : 4x^2 + y^2 \leq 8 \text{ and } y^2 \leq 4x\}$ is :
If $y = y(x)$ satisfies the differential equation
$16(\sqrt{x+ 9\sqrt{x}})(4 + \sqrt{9 + \sqrt{x}}) \cos y \, dy = (1 + 2 \sin y) dx, x > 0$ and $y(256) = \frac{\pi}{2}, y(49) = \alpha$, then $2 \sin \alpha$ is equal to :- Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f\left(\frac{tx}{36}\right) dt = 4\alpha f(x)$. If $y = f(x)$ is a standard parabola passing through the points (2, 1) and (– 4, $\beta$), then $\beta^\alpha$ is equal to ______.
