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
- 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 ______.
- Let $[\cdot]$ denote the greatest integer function and $f(x) = \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right]$. Then $12 \sum_{j=1}^\infty f(j)$ is equal to __________.
Let $f$ be a differentiable function satisfying $f(x) = 1 - 2x + \int_{0}^{x} e^{(x - t)} f(t) dt$, $x \in \mathbf{R}$ and let $g(x) = \int_{0}^{x} (f(t) + 2)^{15} (t - 4)^{6} (t + 12)^{17} dt$, $x \in \mathbf{R}$. If $\text{p}$ and $\text{q}$ are respectively the points of local minima and local maxima of $g$, then the value of $|\text{p} + \text{q}|$ is equal to _________.
Let $f$ be a twice differentiable function such that $f(x) = \int_{0}^{x} \tan(t - x)dt - \int_{0}^{x} f(t)\tan t dt$, $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
Then $f''\left(\frac{\pi}{6}\right) + 12f'\left(-\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)$ is equal to __________- Let $y = y(x)$ be the solution of the differential equation $\left(x^2 - x\sqrt{x^2 - 1}\right)dy + \left(y\left(x - \sqrt{x^2 - 1}\right) - x\right)dx = 0, x \geq 1$. If $y(1) = 1$, then the greatest integer less than $y\left(\sqrt{5}\right)$ is _________.
- If $I_1 = \int_0^1 e^{-x} cos^2x dx$, $I_2 = \int_0^1 e^{-x^2} cos^2x dx$ and $I_3 = \int_0^1 e^{-x^2} dx$; then :
- If $\int \frac{(\sqrt{1+x^2}+x)^{10}}{(\sqrt{1+x^2}-x)^9} dx = \frac{1}{m} \left( (\sqrt{1+x^2}+x)^n (n\sqrt{1+x^2}-x) \right) + C$ where $C$ is the constant of integration and $m, n \in N$, then $m+n$ is equal to
Let [.] denote the greatest integer function. If $\int_{0}^{e^3} \left[\frac{1}{e^{x-1}}\right] dx = \alpha - \log_e 2$, then $\alpha^3$ is equal to ____________.
Let $f: R\to R$ be a thrice differentiable odd function satisfying $f'(x)\ge0, f''(x)=f(x), f(0)=0, f'(0)=3$. Then $9f(\log_e 3)$ is equal to ___________.
- Let $y = y (x)$ be the solution curve of the differentialequation $x (x^2 + e^x) dy + (e^x (x-2) y-x^3) dx = 0, x > 0$, passing through the point $(1, 0)$.Then $y (2)$ is equal to
Important Questions from Integral Calculus
- 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 ______.
- Let $[\cdot]$ denote the greatest integer function and $f(x) = \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right]$. Then $12 \sum_{j=1}^\infty f(j)$ is equal to __________.
Let $f$ be a differentiable function satisfying $f(x) = 1 - 2x + \int_{0}^{x} e^{(x - t)} f(t) dt$, $x \in \mathbf{R}$ and let $g(x) = \int_{0}^{x} (f(t) + 2)^{15} (t - 4)^{6} (t + 12)^{17} dt$, $x \in \mathbf{R}$. If $\text{p}$ and $\text{q}$ are respectively the points of local minima and local maxima of $g$, then the value of $|\text{p} + \text{q}|$ is equal to _________.
Let $f$ be a twice differentiable function such that $f(x) = \int_{0}^{x} \tan(t - x)dt - \int_{0}^{x} f(t)\tan t dt$, $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
Then $f''\left(\frac{\pi}{6}\right) + 12f'\left(-\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)$ is equal to __________- Let $y = y(x)$ be the solution of the differential equation $\left(x^2 - x\sqrt{x^2 - 1}\right)dy + \left(y\left(x - \sqrt{x^2 - 1}\right) - x\right)dx = 0, x \geq 1$. If $y(1) = 1$, then the greatest integer less than $y\left(\sqrt{5}\right)$ is _________.
