Basic Mathematics

Given acceleration $a = kt$.
Since $a = \frac{dv}{dt}$, we can write:
$\frac{dv}{dt} = kt$
$dv = kt dt$
Integrating both sides with initial conditions $v = 0$ at $t = 0$ (motion starts from rest):
$\int_0^v dv = \int_0^t kt dt$
$\vert{}v\vert{}_0^v = k \left\vert{} \frac{t^2}{2} \right\vert{}_0^t$
$v - 0 = k \left( \frac{t^2}{2} - 0 \right)$
$v = \frac{kt^2}{2}$

Given velocity $v(t) = v_0 e^{-\lambda t}$.
First, find the initial time $t_1$ when velocity is $v_0$:
$v(t_1) = v_0 \implies v_0 e^{-\lambda t_1} = v_0 \implies e^{-\lambda t_1} = 1 \implies t_1 = 0$
Next, find the final time $t_2$ when velocity decreases to $\frac{v_0}{2}$:
$v(t_2) = \frac{v_0}{2} \implies v_0 e^{-\lambda t_2} = \frac{v_0}{2}$
$e^{-\lambda t_2} = \frac{1}{2} = 2^{-1}$
Taking natural logarithm on both sides:
$-\lambda t_2 = -\log_e 2 \implies t_2 = \frac{\log_e 2}{\lambda}$
The duration of the time interval is:
$\Delta t = t_2 - t_1 = \frac{\log_e 2}{\lambda} - 0 = \frac{\log_e 2}{\lambda}$
Now, evaluate the displacement $\int_{t_1}^{t_2} v dt$:
$\int_0^{\frac{\log_e 2}{\lambda}} v_0 e^{-\lambda t} dt = v_0 \left\vert{} \frac{e^{-\lambda t}}{-\lambda} \right\vert{}_0^{\frac{\log_e 2}{\lambda}}$
$= -\frac{v_0}{\lambda} \left( e^{-\lambda \left(\frac{\log_e 2}{\lambda}\right)} - e^0 \right)$
$= -\frac{v_0}{\lambda} \left( e^{-\log_e 2} - 1 \right) = -\frac{v_0}{\lambda} \left( \frac{1}{2} - 1 \right) = \frac{v_0}{2\lambda}$
Calculating average velocity $\bar{v}$:
$\bar{v} = \frac{\int_{t_1}^{t_2} v dt}{t_2 - t_1} = \frac{\frac{v_0}{2\lambda}}{\frac{\log_e 2}{\lambda}} = \frac{v_0}{2 \log_e 2}$

Given equations of the curves are: $x^2 + y^2 = 32$ $y = x$ First, find the points of intersection of the line and the circle in the first quadrant. Substituting $y = x$ into $x^2 + y^2 = 32$: $x^2 + x^2 = 32 \implies 2x^2 = 32 \implies x^2 = 16 \implies x = \pm 4$ Since we are considering the first quadrant, $x = 4$ and $y = 4$. So the line and circle intersect at $(4, 4)$. The circle meets the X-axis at $x = \sqrt{32} = 4\sqrt{2}$. The required area $A$ of region $OABO$ is divided into two parts: $A = \text{Area of } ODBO + \text{Area of } DAB D$ $A = \int_0^4 y_{\text{line}} dx + \int_4^{4\sqrt{2}} y_{\text{circle}} dx$ $A = \int_0^4 x dx + \int_4^{4\sqrt{2}} \sqrt{32 - x^2} dx$ Evaluating the first integral: $\int_0^4 x dx = \left\vert{} \frac{x^2}{2} \right\vert{}_0^4 = \frac{16}{2} - 0 = 8$ Evaluating the second integral using $\int \sqrt{a^2 - x^2} dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right)$: $\int_4^{4\sqrt{2}} \sqrt{(4\sqrt{2})^2 - x^2} dx = \left\vert{} \frac{x}{2} \sqrt{32 - x^2} + 16 \sin^{-1}\left(\frac{x}{4\sqrt{2}}\right) \right\vert{}_4^{4\sqrt{2}}$ Substitute upper limit $x = 4\sqrt{2}$: $= \frac{4\sqrt{2}}{2} \sqrt{32 - 32} + 16 \sin^{-1}(1) = 0 + 16 \left(\frac{\pi}{2}\right) = 8\pi$ Substitute lower limit $x = 4$: $= \frac{4}{2} \sqrt{32 - 16} + 16 \sin^{-1}\left(\frac{4}{4\sqrt{2}}\right) = 2(4) + 16 \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = 8 + 16\left(\frac{\pi}{4}\right) = 8 + 4\pi$ So the second integral value is: $8\pi - (8 + 4\pi) = 4\pi - 8$ Combining both parts: $A = 8 + (4\pi - 8) = 4\pi \text{ sq units}$

Given curve is a parabola $y^2 = 4x$, which gives $y = 2\sqrt{x}$ for the upper branch. The line is $x = 4$. The bounded region $OACBO$ is symmetrical about the X-axis. Thus, the total area $A$ is: $A = 2 \times (\text{Area of region } OACO)$ $A = 2 \int_0^4 y_{\text{parabola}} dx$ $A = 2 \int_0^4 2\sqrt{x} dx$ $A = 4 \int_0^4 x^{1/2} dx$ Integrating using the power rule $\int x^n dx = \frac{x^{n+1}}{n+1}$: $A = 4 \left\vert{} \frac{x^{3/2}}{3/2} \right\vert{}_0^4$ $A = 4 \cdot \frac{2}{3} \left\vert{} x^{3/2} \right\vert{}_0^4$ $A = \frac{8}{3} \left( 4^{3/2} - 0 \right)$ Since $4^{3/2} = (2^2)^{3/2} = 2^3 = 8$: $A = \frac{8}{3} \times 8 = \frac{64}{3} \text{ sq units}$

Given lines are $y = 3x + 2$, $y = 0$ (X-axis), $x = -1$, and $x = 1$. First, find the point where $y = 3x + 2$ intersects the X-axis ($y = 0$): $3x + 2 = 0 \implies x = -\frac{2}{3}$ For $x \in \left[-1, -\frac{2}{3}\right]$, $y$ is negative, so $y \le 0$. For $x \in \left[-\frac{2}{3}, 1\right]$, $y$ is positive, so $y \ge 0$. Therefore, the total area $A$ is divided into two regions $EFDE$ and $ABDA$: $A = \int_{-1}^{-2/3} (-y) dx + \int_{-2/3}^1 y dx$ $A = -\int_{-1}^{-2/3} (3x + 2) dx + \int_{-2/3}^1 (3x + 2) dx$ Evaluating the antiderivative $\int (3x + 2) dx = \frac{3x^2}{2} + 2x$: For the first integral from $-1$ to $-\frac{2}{3}$: $\left\vert{} \frac{3x^2}{2} + 2x \right\vert{}_{-1}^{-2/3} = \left( \frac{3\left(-\frac{2}{3}\right)^2}{2} + 2\left(-\frac{2}{3}\right) \right) - \left( \frac{3(-1)^2}{2} + 2(-1) \right)$ $= \left( \frac{3 \cdot \frac{4}{9}}{2} - \frac{4}{3} \right) - \left( \frac{3}{2} - 2 \right)$ $= \left( \frac{2}{3} - \frac{4}{3} \right) - \left( -\frac{1}{2} \right) = -\frac{2}{3} + \frac{1}{2} = -\frac{1}{6}$ Since this region lies below the X-axis, the magnitude of area is: $\text{Area}_1 = -\left(-\frac{1}{6}\right) = \frac{1}{6}$ For the second integral from $-\frac{2}{3}$ to $1$: $\left\vert{} \frac{3x^2}{2} + 2x \right\vert{}_{-2/3}^1 = \left( \frac{3(1)^2}{2} + 2(1) \right) - \left( -\frac{2}{3} \right)$ $= \left( \frac{3}{2} + 2 \right) - \left( -\frac{2}{3} \right) = \frac{7}{2} + \frac{2}{3} = \frac{25}{6}$ Combining both areas: $A = \frac{1}{6} + \frac{25}{6} = \frac{26}{6} = \frac{13}{3} \text{ sq units}$

Given equations of the curves are:
$y = x^3$
$y = x + 6$
$x = 0$ (Y-axis)
First, find the point of intersection between $y = x^3$ and $y = x + 6$:
$x^3 = x + 6 \implies x^3 - x - 6 = 0$
By inspection, testing $x = 2$:
$2^3 - 2 - 6 = 8 - 8 = 0$
Thus, $x = 2$ is the point of intersection in the first quadrant.
For the region between $x = 0$ and $x = 2$, the line $y = x + 6$ lies above the curve $y = x^3$.
So, the bounded area $A$ is given by:
$A = \int_0^2 [(x + 6) - x^3] dx$
Integrating term by term:
$A = \left\vert{} \frac{x^2}{2} + 6x - \frac{x^4}{4} \right\vert{}_0^2$
Substituting the upper limit $x = 2$ and lower limit $x = 0$:
$A = \left( \frac{2^2}{2} + 6(2) - \frac{2^4}{4} \right) - (0)$
$A = \left( \frac{4}{2} + 12 - \frac{16}{4} \right)$
$A = 2 + 12 - 4$
$A = 10 \text{ sq units}$

Given equations:
Parabola: $y = \frac{3x^2}{4}$
Line: $3x - 2y + 12 = 0 \implies y = \frac{3x + 12}{2}$
Find the points of intersection by equating the two expressions for $y$:
$\frac{3x^2}{4} = \frac{3x + 12}{2}$
Multiply both sides by 4:
$3x^2 = 2(3x + 12)$
$3x^2 = 6x + 24$
$3x^2 - 6x - 24 = 0$
Divide by 3:
$x^2 - 2x - 8 = 0$
$(x - 4)(x + 2) = 0$
$x = -2$ and $x = 4$
The upper curve in the interval $[-2, 4]$ is the line $y = \frac{3x + 12}{2}$, and the lower curve is the parabola $y = \frac{3x^2}{4}$.
Set up the area integral:
$A = \int_{-2}^{4} \left( \frac{3x + 12}{2} - \frac{3x^2}{4} \right) dx$
$A = \frac{1}{4} \int_{-2}^{4} (6x + 24 - 3x^2) dx$
$A = \frac{1}{4} \left[ 3x^2 + 24x - x^3 \right]_{-2}^{4}$
Evaluate at upper limit $x = 4$:
$3(4)^2 + 24(4) - (4)^3 = 3(16) + 96 - 64 = 48 + 96 - 64 = 80$
Evaluate at lower limit $x = -2$:
$3(-2)^2 + 24(-2) - (-2)^3 = 3(4) - 48 - (-8) = 12 - 48 + 8 = -28$
Subtract the lower limit evaluation from the upper limit evaluation:
$A = \frac{1}{4} \left( 80 - (-28) \right)$
$A = \frac{1}{4} (108)$
$A = 27 \text{ sq units}$

Given equation of the ellipse:
$\frac{x^2}{4} + \frac{y^2}{9} = 1$
Comparing with the standard equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:
$a^2 = 4 \implies a = 2$
$b^2 = 9 \implies b = 3$
The total area of the ellipse is given by:
$A_{\text{total}} = \pi a b = \pi (2)(3) = 6\pi \text{ sq units}$
The area of the region under the curve and above the $X$-axis represents the upper half of the ellipse:
$A = \frac{1}{2} A_{\text{total}} = \frac{1}{2} (6\pi) = 3\pi \text{ sq units}$