Application Of Integrations

Integration can determine the area bounded by a curve and a line. This area can represent physical quantities like work done or distance traveled.

i) Area above the x-axis

Consider the area bounded by the curve $y = f(x)$, the x-axis, and the lines $x = a$ and $x = b$.

Formula: The area under the curve is given by the definite integral:

$\int_a^b f(x)\, dx = F(x)\Big|_a^b = F(b) - F(a)$

Where $F(x)$ is the antiderivative of $f(x)$.

Example 1: Find the area under the curve $f(x) = x^2 + 1$ from $x = 0$ to $x = 2$.

Solution:

$\int_0^2 (x^2 + 1)\, dx = \left[\frac{x^3}{3} + x\right]_0^2$

$= \left(\frac{2^3}{3} + 2\right) - \left(\frac{0^3}{3} + 0\right)$

$= \left(\frac{8}{3} + 2\right) - 0 = \frac{14}{3}$

Therefore, the area under the curve is $14/3$ square units.

Example 2: Find the area under the curve $f(x) = x^3 + 1$ from $x = -1$ to $x = 0$.

Solution:

$\int_{-1}^0 (x^3 + 1)\, dx = \left[\frac{x^4}{4} + x\right]_{-1}^0$

$= \left(\frac{0^4}{4} + 0\right) - \left(\frac{(-1)^4}{4} + (-1)\right)$

$= 0 - \left(\frac{1}{4} - 1\right) = \frac{3}{4}$

Therefore, the area under the curve is $3/4$ square units.

ii) Area below the x-axis

If the curve is below the x-axis, the integral will give a negative value. Since area is always positive, take the absolute value of the result.

Formula: $A = -\int_a^b f(x)\, dx$ or $A = \left|\int_a^b f(x)\, dx\right|$

Example 1: Find the area under the curve $f(x) = -x^2 + 1$ from $x = 1$ to $x = 2$.

Solution:

$\int_1^2 (-x^2 + 1)\, dx = \left[-\frac{x^3}{3} + x\right]_1^2$

$= \left(-\frac{8}{3} + 2\right) - \left(-\frac{1}{3} + 1\right) = -\frac{2}{3} - \frac{2}{3} = -\frac{4}{3}$

The actual area is $|-4/3| = 4/3$ square units.

Area with parts above and below the x-axis: Calculate the areas above and below the x-axis separately and then add their absolute values.

Example 2: Find the total area enclosed by the curve $f(x) = 4x^2 - 1$ and the x-axis between $x = 0$ and $x = 1$.

Solution: The curve intersects the x-axis at $x = \pm 0.5$.

$\int_0^1 |4x^2 - 1|\, dx = \int_0^{0.5} (1-4x^2)\, dx + \int_{0.5}^1 (4x^2-1)\, dx$

$= \left[x - \frac{4x^3}{3}\right]_0^{0.5} + \left[\frac{4x^3}{3} - x\right]_{0.5}^1$

$= \left(\frac{1}{2} - \frac{1}{6}\right) + \left(\frac{4}{3} - 1 - \left(\frac{1}{6} - \frac{1}{2}\right)\right)$

$= \frac{1}{3} + \frac{2}{3} = 1$

The required area is 1 square units.

iii) Area between two intersecting curves

If two curves $y = f(x)$ and $y = g(x)$ intersect at points A and B ($x=a$ and $x=b$), the area between them is:

Formula: $A = \int_a^b |g(x) - f(x)|\, dx$

Where $g(x)$ is the upper curve and $f(x)$ is the lower curve in the interval $[a, b]$.

Example 1: Find the area bounded by the curve $f(x) = x^2$ and the straight line $g(x) = x$.

Solution: The curves intersect when $x^2 = x$, which gives $x = 0$ and $x = 1$.

$A = \int_0^1 (x - x^2)\, dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1$

$= \left(\frac{1}{2} - \frac{1}{3}\right) - 0 = \frac{1}{6}$

The area is $1/6$ square units.

Example 2: Find the area bounded by the curves $f(x) = x^2$ and $g(x) = 2x - x^2$.

Solution: The curves intersect when $x^2 = 2x - x^2$, which simplifies to $2x^2 - 2x = 0$, giving $x = 0$ and $x = 1$.

$A = \int_0^1 (2x - x^2 - x^2)\, dx = \int_0^1 (2x - 2x^2)\, dx$

$= \left[x^2 - \frac{2x^3}{3}\right]_0^1 = \left(1 - \frac{2}{3}\right) - 0 = \frac{1}{3}$

The area is $1/3$ square units.

A solid of revolution is formed by rotating a region bounded by a curve around an axis.

Rotation about the x-axis: $V = \pi\int_a^b y^2\, dx$

Rotation about the y-axis: $V = \pi\int_a^b x^2\, dy$

Example 1: Find the volume of the solid formed by rotating the region bounded by $y = x$, the x-axis, and the lines $x = 1$ and $x = 3$ about the x-axis.

Solution:

$V = \pi\int_1^3 x^2\, dx = \pi\left[\frac{x^3}{3}\right]_1^3$

$= \pi\left(\frac{27}{3} - \frac{1}{3}\right) = \frac{26\pi}{3} \text{ cubic units}$

Example 2: Find the volume of the solid generated when the region between the curve $y = 4 - x^2$ and the lines $y = 0$ and $y = 4$ is rotated about the y-axis.

Solution: $x^2 = 4-y$

$V = \pi\int_0^4 (4 - y)\, dy = \pi\left[4y - \frac{y^2}{2}\right]_0^4$

$= \pi(16 - 8) = 8\pi \text{ cubic units}$

Example 3: Find the volume of the solid generated by the curve $xy = 2$ rotating about the x-axis, and the straight lines $x = -4$ and $x = -2$.

Solution: $y = 2/x$

$V = \pi\int_{-4}^{-2} \left(\frac{2}{x}\right)^2 dx = 4\pi\int_{-4}^{-2} \frac{1}{x^2}\, dx$

$= 4\pi\left[-\frac{1}{x}\right]_{-4}^{-2} = 4\pi\left(\frac{1}{2} - \frac{1}{4}\right) = \pi \text{ cubic units}$

1. Marginal Cost (MC): The additional cost of producing one more unit.
2. Marginal Revenue (MR): The additional revenue from selling one more unit.
3. Total Cost ($C(x)$): $C(x) = \int MC(x)\, dx + \text{Fixed Cost}$
4. Total Revenue ($R(x)$): $R(x) = \int MR(x)\, dx$

Example 1: The marginal cost of a firm is given by $M(x) = 20 + 0.5x^2 - 3x^3$. Find the total cost function.

Solution:

$C(x) = \int (20 + 0.5x^2 - 3x^3)\, dx$

$= 20x + \frac{0.5x^3}{3} - \frac{3x^4}{4} + c$

$= 20x + \frac{1}{6}x^3 - \frac{3}{4}x^4 + c$

Example 2: Given that the fixed cost is 30 and the marginal cost, $MC(x) = 3x - \frac{1}{2}x^2$. Determine the total cost function.

Solution:

$C(x) = \int \left(3x - \frac{1}{2}x^2\right) dx = \frac{3}{2}x^2 - \frac{1}{6}x^3 + c$

Since $C(x) = FC + VC(x)$ where $FC$ represents fixed cost and $VC$ represents variable cost.

Therefore, $C(x) = 30 + \frac{3}{2}x^2 - \frac{1}{6}x^3$

Example 3: Assume that the marginal cost in Tanzanian shillings is given by $MC(x) = \frac{3}{7}e^{-2x} + 6x^2 + 11$, where $x$ is the quantity produced. Find the total cost of production when $x = 3$, if the fixed cost is Tsh 10,000.

Solution:

$C(x) = \int \left(\frac{3}{7}e^{-2x} + 6x^2 + 11\right) dx$

$= -\frac{3}{14}e^{-2x} + 2x^3 + 11x + T$

When $x = 0$, $C(x) = 10000$:

$10000 = -\frac{3}{14} + T$

$T \approx 10000.21$

So, $C(x) = -\frac{3}{14}e^{-2x} + 2x^3 + 11x + 10000.21$

$C(3) = -\frac{3}{14}e^{-6} + 2(3)^3 + 11(3) + 10000.21 \approx 10087.21 \text{ Tsh}$

Integration is crucial in describing motion using vector quantities like displacement, velocity, and acceleration.

Velocity ($v(t)$): $v(t) = \int a(t)\, dt$ (integral of acceleration)

Displacement ($s(t)$): $s(t) = \int v(t)\, dt$ (integral of velocity)

Example 1: A particle is moving along a straight line with a velocity of $V(t) = t^3 - 4t^2 - 4t + 16$ meters per second at time $t$. What is the displacement of the particle within the time interval $1 \le t \le 5$?

Solution:

$s(t) = \int_1^5 (t^3 - 4t^2 - 4t + 16)\, dt$

$= \left[\frac{t^4}{4} - \frac{4t^3}{3} - 2t^2 + 16t\right]_1^5$

$= \left(\frac{625}{4} - \frac{500}{3} - 50 + 80\right) - \left(\frac{1}{4} - \frac{4}{3} - 2 + 16\right)$

$= \frac{20}{3} \text{ meters}$

Example 2: A car starts from rest and moves with a displacement (in meters) of $s(t) = \frac{1}{6}t^4 + \frac{1}{4}t^3$. What is the velocity at the end of 2 seconds?

Solution:

$v(t) = \frac{d}{dt} s(t) = \frac{d}{dt}\left(\frac{1}{6}t^4 + \frac{1}{4}t^3\right)$

$v(t) = \frac{4}{6}t^3 + \frac{3}{4}t^2 = \frac{2}{3}t^3 + \frac{3}{4}t^2$

$v(2) = \frac{2}{3}(2)^3 + \frac{3}{4}(2)^2 = \frac{16}{3} + 3 = \frac{25}{3} = 8\frac{1}{3} \text{ m/s}$

Therefore, the velocity after 2 seconds is $8\frac{1}{3}$ m/s.