Area bounded by region of single curve Questions in English

(D) The area $A$ is given by $\int_{0}^{4\pi} x |\sin(\frac{x}{2})| dx$.
Since $\sin(\frac{x}{2}) \ge 0$ for $x \in [0, 2\pi]$ and $\sin(\frac{x}{2}) \le 0$ for $x \in [2\pi, 4\pi]$, we have:
$A = \int_{0}^{2\pi} x \sin(\frac{x}{2}) dx + \int_{2\pi}^{4\pi} x (-\sin(\frac{x}{2})) dx$.
Using integration by parts $\int u dv = uv - \int v du$ with $u=x, dv=\sin(\frac{x}{2})dx$:
$\int x \sin(\frac{x}{2}) dx = -2x \cos(\frac{x}{2}) + 4 \sin(\frac{x}{2})$.
Evaluating the first integral: $[-2x \cos(\frac{x}{2}) + 4 \sin(\frac{x}{2})]_{0}^{2\pi} = (-2(2\pi)(-1) + 0) - (0) = 4\pi$.
Evaluating the second integral: $-[-2x \cos(\frac{x}{2}) + 4 \sin(\frac{x}{2})]_{2\pi}^{4\pi} = -[(-2(4\pi)(1) + 0) - (-2(2\pi)(-1) + 0)] = -[-8\pi - 4\pi] = 12\pi$.
Total area $A = 4\pi + 12\pi = 16\pi$ square units.

(C) The given equation is $y = 2\sqrt{1 - x^2}$.
Squaring both sides, we get $y^2 = 4(1 - x^2)$, which implies $x^2 + \frac{y^2}{4} = 1$.
This is the equation of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a = 1$ and $b = 2$.
The curve $y = 2\sqrt{1 - x^2}$ represents the upper half of the ellipse.
The area of the full ellipse is $A = \pi ab = \pi(1)(2) = 2\pi$.
The area bounded by the curve and the $X$-axis is the area of the upper half of the ellipse.
Area $= \frac{1}{2} \times 2\pi = \pi \text{ sq. units}$.

(B) The region is bounded by $y = x^3$, $y = 8$, and $x = 0$.
Expressing $x$ in terms of $y$, we get $x = y^{1/3}$.
The intersection of $y = x^3$ and $y = 8$ occurs at $x^3 = 8$, which gives $x = 2$.
The area $A$ is given by the integral of $x$ with respect to $y$ from $y = 0$ to $y = 8$:
$A = \int_{0}^{8} y^{1/3} \, dy$
$A = \left[ \frac{y^{4/3}}{4/3} \right]_{0}^{8}$
$A = \frac{3}{4} [8^{4/3} - 0^{4/3}]$
$A = \frac{3}{4} [(2^3)^{4/3}] = \frac{3}{4} [2^4] = \frac{3}{4} \times 16 = 12$ square units.

(C) The area $A$ is given by the integral $\int_{0}^{2} 2^{kx} dx = \frac{2}{\ln 2}$.
Using the formula $\int a^x dx = \frac{a^x}{\ln a}$, we have $\int_{0}^{2} 2^{kx} dx = \left[ \frac{2^{kx}}{k \ln 2} \right]_{0}^{2}$.
Substituting the limits: $\frac{1}{k \ln 2} (2^{2k} - 2^0) = \frac{2}{\ln 2}$.
$\frac{2^{2k} - 1}{k} = 2$.
$2^{2k} - 1 = 2k$.
By inspection, if $k = 1$, $2^2 - 1 = 3 \neq 2(1)$.
If $k = 1/2$, $2^{2(1/2)} - 1 = 2^1 - 1 = 1$, and $2k = 2(1/2) = 1$.
Since $1 = 1$, the value of $k$ is $\frac{1}{2}$.

(A) The equation of the curve is $x^2 = by$, which implies $x = \sqrt{b} \sqrt{y}$ (since it is in the first quadrant).
The area $A$ is given by the integral of $x$ with respect to $y$ from $y = 1$ to $y = 4$:
$A = \int_{1}^{4} \sqrt{b} \sqrt{y} \, dy = 28$
$\sqrt{b} \int_{1}^{4} y^{1/2} \, dy = 28$
$\sqrt{b} \left[ \frac{y^{3/2}}{3/2} \right]_{1}^{4} = 28$
$\sqrt{b} \cdot \frac{2}{3} [4^{3/2} - 1^{3/2}] = 28$
$\sqrt{b} \cdot \frac{2}{3} [8 - 1] = 28$
$\sqrt{b} \cdot \frac{2}{3} \cdot 7 = 28$
$\sqrt{b} = \frac{28 \cdot 3}{14} = 2 \cdot 3 = 6$
$b = 6^2 = 36$.

(D) The area $A$ is given by the integral $\int_{3}^{5} |x - 4| \, dx$.
Since $|x - 4| = -(x - 4)$ for $x < 4$ and $|x - 4| = (x - 4)$ for $x \ge 4$, we split the integral:
$A = \int_{3}^{4} -(x - 4) \, dx + \int_{4}^{5} (x - 4) \, dx$.
Evaluating the first part: $\int_{3}^{4} (-x + 4) \, dx = [-\frac{x^2}{2} + 4x]_{3}^{4} = (-8 + 16) - (-4.5 + 12) = 8 - 7.5 = 0.5$.
Evaluating the second part: $\int_{4}^{5} (x - 4) \, dx = [\frac{x^2}{2} - 4x]_{4}^{5} = (12.5 - 20) - (8 - 16) = -7.5 - (-8) = 0.5$.
Total area $A = 0.5 + 0.5 = 1$ sq. unit.

(A) Step $1$: Express $x$ in terms of $y$ from the equation $2x - y + 1 = 0$. We get $2x = y - 1$, so $x = \frac{y - 1}{2}$.
Step $2$: The area $A$ bounded by the curve $x = f(y)$, the $y$-axis, and the lines $y = c$ and $y = d$ is given by $A = \int_{c}^{d} |x| \, dy$.
Step $3$: Substitute the limits $y = -1$ to $y = 3$: $A = \int_{-1}^{3} |\frac{y - 1}{2}| \, dy$.
Step $4$: The expression $\frac{y - 1}{2}$ changes sign at $y = 1$. Thus, $A = \int_{-1}^{1} -(\frac{y - 1}{2}) \, dy + \int_{1}^{3} (\frac{y - 1}{2}) \, dy$.
Step $5$: Calculate the integrals: $\frac{1}{2} [-( \frac{y^2}{2} - y )]_{-1}^{1} + \frac{1}{2} [ \frac{y^2}{2} - y ]_{1}^{3} = \frac{1}{2} [-( (\frac{1}{2} - 1) - (\frac{1}{2} + 1) )] + \frac{1}{2} [ (\frac{9}{2} - 3) - (\frac{1}{2} - 1) ] = \frac{1}{2} [ -(-\frac{1}{2} - \frac{3}{2}) ] + \frac{1}{2} [ \frac{3}{2} - (-\frac{1}{2}) ] = \frac{1}{2} [2] + \frac{1}{2} [2] = 1 + 1 = 2$ square units.

(D) The area $A$ is given by the integral of $|y|$ with respect to $x$ from $x = -2$ to $x = 4$.
$A = \int_{-2}^{4} |x| \, dx$
Since $|x| = -x$ for $x < 0$ and $|x| = x$ for $x \ge 0$, we split the integral:
$A = \int_{-2}^{0} (-x) \, dx + \int_{0}^{4} x \, dx$
$A = \left[ -\frac{x^2}{2} \right]_{-2}^{0} + \left[ \frac{x^2}{2} \right]_{0}^{4}$
$A = (0 - (-\frac{(-2)^2}{2})) + (\frac{4^2}{2} - 0)$
$A = (0 - (-2)) + (8 - 0)$
$A = 2 + 8 = 10 \text{ square units}$.

(B) The area $A$ is given by $\int_{-2}^{1} |x^3 + ax| \, dx$.
Since $y = x^3 + ax = x(x^2 + a)$, and $a > 0$, the curve crosses the $x$-axis only at $x = 0$ in the interval $[-2, 1]$.
For $x \in [-2, 0]$, $x^3 + ax \le 0$, so $|x^3 + ax| = -(x^3 + ax)$.
For $x \in [0, 1]$, $x^3 + ax \ge 0$, so $|x^3 + ax| = x^3 + ax$.
Area $= \int_{-2}^{0} -(x^3 + ax) \, dx + \int_{0}^{1} (x^3 + ax) \, dx = \frac{37}{4}$.
Evaluating the integrals: $-[\frac{x^4}{4} + \frac{ax^2}{2}]_{-2}^{0} + [\frac{x^4}{4} + \frac{ax^2}{2}]_{0}^{1} = \frac{37}{4}$.
$-(0 - (4 + 2a)) + (\frac{1}{4} + \frac{a}{2}) = \frac{37}{4}$.
$4 + 2a + \frac{1}{4} + \frac{a}{2} = \frac{37}{4}$.
$\frac{5a}{2} = \frac{37}{4} - \frac{17}{4} = \frac{20}{4} = 5$.
$a = 2$.

(A) Step $1$: Find the points of intersection with the $X$-axis by setting $y = 0$.
$2x - x^2 = 0 \implies x(2 - x) = 0$.
So, $x = 0$ and $x = 2$.
Step $2$: The area $A$ is given by the integral $\int_{0}^{2} (2x - x^2) \, dx$.
Step $3$: Evaluate the integral: $\left[ x^2 - \frac{x^3}{3} \right]_{0}^{2}$.
Step $4$: Substitute the limits: $\left( 2^2 - \frac{2^3}{3} \right) - (0 - 0) = 4 - \frac{8}{3} = \frac{12 - 8}{3} = \frac{4}{3} \text{ sq. units}$.

(A) Given the curve $y = 4x^2$, we express $x$ in terms of $y$ as $x = \sqrt{\frac{y}{4}} = \frac{\sqrt{y}}{2}$.
The area $A$ in the first quadrant bounded by the $Y$-axis and the lines $y = 2$ and $y = 4$ is given by the integral:
$A = \int_{2}^{4} x \, dy = \int_{2}^{4} \frac{\sqrt{y}}{2} \, dy$.
$A = \frac{1}{2} \int_{2}^{4} y^{1/2} \, dy = \frac{1}{2} \left[ \frac{y^{3/2}}{3/2} \right]_{2}^{4} = \frac{1}{2} \cdot \frac{2}{3} [y^{3/2}]_{2}^{4}$.
$A = \frac{1}{3} [4^{3/2} - 2^{3/2}] = \frac{1}{3} [8 - 2\sqrt{2}]$ sq. units.

(D) The given curve is $y = \sqrt{16 - x^2}$, which represents the upper semi-circle of $x^2 + y^2 = 4^2$ with radius $r = 4$.
We need to find the area bounded by $x = 0$ and $x = 4$ above the $X$-axis.
This region represents one-quarter of the circle with radius $r = 4$.
The area of a full circle is $A = \pi r^2 = \pi(4)^2 = 16\pi$.
The area of the required region is $\frac{1}{4} \times 16\pi = 4\pi$ square units.

(B) Given the curve $y^2 = x^3$, we express $x$ in terms of $y$ as $x = y^{2/3}$.
The area $A$ bounded by the curve, the $y$-axis, and the lines $y = 1$ and $y = 8$ is given by the integral:
$A = \int_{1}^{8} x \, dy = \int_{1}^{8} y^{2/3} \, dy$
Integrating $y^{2/3}$ with respect to $y$:
$A = \left[ \frac{y^{(2/3) + 1}}{(2/3) + 1} \right]_{1}^{8} = \left[ \frac{y^{5/3}}{5/3} \right]_{1}^{8} = \left[ \frac{3}{5} y^{5/3} \right]_{1}^{8}$
Evaluating the definite integral:
$A = \frac{3}{5} [8^{5/3} - 1^{5/3}]$
Since $8^{5/3} = (2^3)^{5/3} = 2^5 = 32$ and $1^{5/3} = 1$:
$A = \frac{3}{5} [32 - 1] = \frac{3}{5} \times 31 = \frac{93}{5} \text{ sq. units}$.