4.5 Application Of Antiderivatives

Detailed Theory: Applications of Antiderivatives

1. Introduction to Applications of Antiderivatives

1.1 What are Antiderivatives?

An antiderivative of a function $f(x)$ is a function $F(x)$ such that:

$$F'(x) = f(x)$$

The process of finding antiderivatives is called integration or indefinite integration.

1.2 Why Study Applications?

Antiderivatives have powerful real-world applications in:

• Finding area under curves

• Computing accumulated quantities

• Solving differential equations

• Physics: displacement from velocity, velocity from acceleration

• Economics: total cost from marginal cost

1.3 Notation

• Indefinite integral: $\int f(x) dx = F(x) + C$

• Definite integral: $\int_{a}^{b} f(x) dx = F(b) - F(a)$

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2. Area Under a Curve

2.1 Basic Concept

The definite integral $\int_{a}^{b} f(x) dx$ represents the net area between the curve $y = f(x)$ and the x-axis from $x = a$ to $x = b$.

2.2 Net Area vs Total Area

• Net area: $\int_{a}^{b} f(x) dx$ (areas below x-axis are negative)

• Total area: $\int_{a}^{b} |f(x)| dx$ (all areas are positive)

2.3 Computing Area

Step-by-step:

1. Sketch the curve to identify regions above/below x-axis

2. Find x-intercepts within interval [a, b]

3. Split integral at intercepts

4. Integrate each piece

5. Take absolute value for total area

2.4 Example 1: Net Area

Find net area between $y = x^2 - 1$ and x-axis from $x = 0$ to $x = 2$

Solution: x-intercepts: $x^2 - 1 = 0 \Rightarrow x = \pm 1$

Within [0, 2]: intercept at $x = 1$

Split: $\int_{0}^{2} (x^2 - 1) dx = \int_{0}^{1} (x^2 - 1) dx + \int_{1}^{2} (x^2 - 1) dx$

First integral (0 to 1, below x-axis):

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

Second integral (1 to 2, above x-axis):

$$\int_{1}^{2} (x^2 - 1) dx = \left[\frac{x^3}{3} - x\right]_{1}^{2} = \left(\frac{8}{3} - 2\right) - \left(\frac{1}{3} - 1\right) = \frac{8}{3} - 2 - \frac{1}{3} + 1$$

$$= \frac{7}{3} - 1 = \frac{4}{3}$$

Net area: $-\frac{2}{3} + \frac{4}{3} = \frac{2}{3}$

2.5 Example 2: Total Area

Find total area between $y = \sin x$ and x-axis from $x = 0$ to $x = 2\pi$

Solution: x-intercepts in [0, 2π]: $x = 0, \pi, 2\pi$

Total area = $\int_{0}^{\pi} \sin x dx + \int_{\pi}^{2\pi} (-\sin x) dx$

First: $\int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi} = (-\cos \pi) - (-\cos 0) = 1 + 1 = 2$

Second: $\int_{\pi}^{2\pi} (-\sin x) dx = [\cos x]_{\pi}^{2\pi} = \cos 2\pi - \cos \pi = 1 - (-1) = 2$

Total area = 2 + 2 = 4

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3. Area Between Two Curves

3.1 General Formula

Area between $y = f(x)$ (upper) and $y = g(x)$ (lower) from $x = a$ to $x = b$:

$$A = \int_{a}^{b} [f(x) - g(x)] dx$$

Important: $f(x) \geq g(x)$ on [a, b]

3.2 Steps for Finding Area Between Curves

1. Find intersection points (solve $f(x) = g(x)$)

2. Determine which function is above on each interval

3. Set up and evaluate integrals

3.3 Example 1: Simple Case

Find area between $y = x^2$ and $y = x$ from $x = 0$ to $x = 1$

Solution: Intersection: $x^2 = x \Rightarrow x(x-1) = 0 \Rightarrow x = 0, 1$

Between 0 and 1: $x \geq x^2$

$$A = \int_{0}^{1} (x - x^2) dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$$

3.4 Example 2: Switching Order

Find area between $y = x^3$ and $y = x$ from $x = -1$ to $x = 1$

Solution: Intersection: $x^3 = x \Rightarrow x(x^2-1) = 0 \Rightarrow x = -1, 0, 1$

Region 1 ([-1, 0]): $x \geq x^3$ Region 2 ([0, 1]): $x^3 \geq x$

So area = $\int_{-1}^{0} (x - x^3) dx + \int_{0}^{1} (x^3 - x) dx$

First: $\left[\frac{x^2}{2} - \frac{x^4}{4}\right]_{-1}^{0} = 0 - \left(\frac{1}{2} - \frac{1}{4}\right) = -\frac{1}{4}$ (but area is positive)

Second: $\left[\frac{x^4}{4} - \frac{x^2}{2}\right]_{0}^{1} = \left(\frac{1}{4} - \frac{1}{2}\right) - 0 = -\frac{1}{4}$

Wait! Both negative? That means we guessed wrong about which is above.

Actually: Check at $x = 0.5$: $x^3 = 0.125$, $x = 0.5$, so $x > x^3$

So correction: $x \geq x^3$ on entire [-1, 1]

$$A = \int_{-1}^{1} (x - x^3) dx$$

Since integrand is odd ($x - x^3$ is odd), area = 0?

No! Area is positive, we need absolute value.

Better: $A = \int_{-1}^{1} |x - x^3| dx$

From symmetry: $A = 2\int_{0}^{1} (x - x^3) dx = 2\left[\frac{x^2}{2} - \frac{x^4}{4}\right]_{0}^{1} = 2\left(\frac{1}{2} - \frac{1}{4}\right) = 2 \cdot \frac{1}{4} = \frac{1}{2}$

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4. Volume of Solids of Revolution

4.1 Disk Method (Rotation around x-axis)

When region under $y = f(x)$ is rotated about x-axis:

Volume = $\pi \int_{a}^{b} [f(x)]^2 dx$

4.2 Washer Method

When region between $y = f(x)$ (outer) and $y = g(x)$ (inner) is rotated about x-axis:

Volume = $\pi \int_{a}^{b} \left([f(x)]^2 - [g(x)]^2\right) dx$

4.3 Shell Method (Rotation around y-axis)

When region between $y = f(x)$ and x-axis is rotated about y-axis:

Volume = $2\pi \int_{a}^{b} x f(x) dx$

4.4 Example 1: Disk Method

Find volume when region under $y = \sqrt{x}$ from $x = 0$ to $x = 4$ is rotated about x-axis

Solution:

$$V = \pi \int_{0}^{4} (\sqrt{x})^2 dx = \pi \int_{0}^{4} x dx = \pi \left[\frac{x^2}{2}\right]_{0}^{4} = \pi \cdot 8 = 8\pi$$

4.5 Example 2: Washer Method

Find volume when region between $y = x$ and $y = x^2$ is rotated about x-axis

Solution: Intersection: $x = x^2 \Rightarrow x(x-1) = 0 \Rightarrow x = 0, 1$

Between 0 and 1: $x \geq x^2$

$$V = \pi \int_{0}^{1} (x^2 - (x^2)^2) dx = \pi \int_{0}^{1} (x^2 - x^4) dx$$

$$= \pi \left[\frac{x^3}{3} - \frac{x^5}{5}\right]_{0}^{1} = \pi \left(\frac{1}{3} - \frac{1}{5}\right) = \pi \cdot \frac{2}{15} = \frac{2\pi}{15}$$

4.6 Example 3: Shell Method

Find volume when region under $y = x^2$ from $x = 0$ to $x = 2$ is rotated about y-axis

Solution:

$$V = 2\pi \int_{0}^{2} x \cdot x^2 dx = 2\pi \int_{0}^{2} x^3 dx = 2\pi \left[\frac{x^4}{4}\right]_{0}^{2}$$

$$= 2\pi \cdot \frac{16}{4} = 2\pi \cdot 4 = 8\pi$$

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5. Arc Length

5.1 Formula for Arc Length

Length of curve $y = f(x)$ from $x = a$ to $x = b$:

$$L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} dx$$

5.2 Example 1: Straight Line

Find length of $y = 2x + 1$ from $x = 0$ to $x = 3$

Solution: $f'(x) = 2$

$$L = \int_{0}^{3} \sqrt{1 + 2^2} dx = \int_{0}^{3} \sqrt{5} dx = \sqrt{5} [x]_{0}^{3} = 3\sqrt{5}$$

Verification: Points (0,1) and (3,7), distance = $\sqrt{(3-0)^2 + (7-1)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$

5.3 Example 2: Parabola

Find length of $y = x^{3/2}$ from $x = 0$ to $x = 4$

Solution: $f'(x) = \frac{3}{2}x^{1/2}$, so $[f'(x)]^2 = \frac{9}{4}x$

$$L = \int_{0}^{4} \sqrt{1 + \frac{9}{4}x} dx$$

Let $u = 1 + \frac{9}{4}x$, $du = \frac{9}{4}dx$

When $x=0$, $u=1$; when $x=4$, $u=10$

$$L = \int_{1}^{10} \sqrt{u} \cdot \frac{4}{9} du = \frac{4}{9} \cdot \frac{2}{3}[u^{3/2}]_{1}^{10} = \frac{8}{27}(10^{3/2} - 1)$$

$$= \frac{8}{27}(10\sqrt{10} - 1)$$

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6. Applications in Physics

6.1 Displacement from Velocity

If $v(t)$ = velocity at time $t$, then displacement from time $a$ to $b$:

$$\text{Displacement} = \int_{a}^{b} v(t) dt$$

6.2 Distance Traveled

Total distance traveled (always positive):

$$\text{Distance} = \int_{a}^{b} |v(t)| dt$$

6.3 Example: Particle Motion

Particle moves with velocity $v(t) = t^2 - 3t + 2$ (m/s). Find:

1. Displacement from $t=0$ to $t=3$

2. Distance traveled from $t=0$ to $t=3$

Solution:

1. Displacement:

$$\int_{0}^{3} (t^2 - 3t + 2) dt = \left[\frac{t^3}{3} - \frac{3t^2}{2} + 2t\right]_{0}^{3}$$

$$= \left(\frac{27}{3} - \frac{27}{2} + 6\right) - 0 = 9 - 13.5 + 6 = 1.5 \text{ m}$$

1. Distance: First find when $v(t)=0$: $t^2 - 3t + 2 = 0 \Rightarrow (t-1)(t-2) = 0 \Rightarrow t=1, 2$

Sign of $v(t)$:

• [0,1]: Test $t=0.5$, $v(0.5)=0.25-1.5+2=0.75>0$

• [1,2]: Test $t=1.5$, $v(1.5)=2.25-4.5+2=-0.25<0$

• [2,3]: Test $t=2.5$, $v(2.5)=6.25-7.5+2=0.75>0$

So distance = $\int_{0}^{1} v(t) dt - \int_{1}^{2} v(t) dt + \int_{2}^{3} v(t) dt$

Compute each:

$$\int_{0}^{1} v(t) dt = \left[\frac{t^3}{3} - \frac{3t^2}{2} + 2t\right]_{0}^{1} = \frac{1}{3} - \frac{3}{2} + 2 = \frac{1}{3} + 0.5 = \frac{5}{6}$$

$$\int_{1}^{2} v(t) dt = \left[\frac{t^3}{3} - \frac{3t^2}{2} + 2t\right]_{1}^{2} = \left(\frac{8}{3} - 6 + 4\right) - \left(\frac{1}{3} - \frac{3}{2} + 2\right)$$

$$= \left(\frac{8}{3} - 2\right) - \left(\frac{1}{3} + 0.5\right) = \left(\frac{2}{3}\right) - \left(\frac{5}{6}\right) = -\frac{1}{6}$$

$$\int_{2}^{3} v(t) dt = \left[\frac{t^3}{3} - \frac{3t^2}{2} + 2t\right]_{2}^{3} = \left(9 - \frac{27}{2} + 6\right) - \left(\frac{8}{3} - 6 + 4\right)$$

$$= \left(15 - 13.5\right) - \left(\frac{8}{3} - 2\right) = 1.5 - \left(\frac{2}{3}\right) = \frac{3}{2} - \frac{2}{3} = \frac{5}{6}$$

Distance = $\frac{5}{6} - (-\frac{1}{6}) + \frac{5}{6} = \frac{5}{6} + \frac{1}{6} + \frac{5}{6} = \frac{11}{6}$ m

6.4 Work Done by Variable Force

Work = $\int_{a}^{b} F(x) dx$ where $F(x)$ is force as function of position

Example: Spring follows Hooke's Law: $F(x) = kx$. Work to stretch from $x_1$ to $x_2$:

$$W = \int_{x_1}^{x_2} kx dx = \frac{k}{2}(x_2^2 - x_1^2)$$

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7. Applications in Economics

7.1 Total Cost from Marginal Cost

If $C'(x)$ = marginal cost (cost to produce one more unit), then total cost to produce $x$ units:

$$C(x) = \int_{0}^{x} C'(t) dt + C_0$$

where $C_0$ = fixed costs

7.2 Total Revenue from Marginal Revenue

If $R'(x)$ = marginal revenue, then total revenue from selling $x$ units:

$$R(x) = \int_{0}^{x} R'(t) dt$$

7.3 Consumer and Producer Surplus

Consumer Surplus:

$$CS = \int_{0}^{q^*} [D(q) - p^*] dq$$

where $D(q)$ = demand function, $p^*$ = equilibrium price, $q^*$ = equilibrium quantity

Producer Surplus:

$$PS = \int_{0}^{q^*} [p^* - S(q)] dq$$

where $S(q)$ = supply function

7.4 Example: Cost and Revenue

Marginal cost: $C'(x) = 2x + 3$ Marginal revenue: $R'(x) = 10 - x$ Fixed costs: $100

Find:

1. Total cost function

2. Total revenue function

3. Profit function

4. Maximum profit

Solution:

1. $C(x) = \int_{0}^{x} (2t + 3) dt + 100 = [t^2 + 3t]_{0}^{x} + 100 = x^2 + 3x + 100$

2. $R(x) = \int_{0}^{x} (10 - t) dt = [10t - \frac{t^2}{2}]_{0}^{x} = 10x - \frac{x^2}{2}$

3. Profit: $P(x) = R(x) - C(x) = \left(10x - \frac{x^2}{2}\right) - (x^2 + 3x + 100)$

$$= 10x - \frac{x^2}{2} - x^2 - 3x - 100 = 7x - \frac{3x^2}{2} - 100$$

1. Maximize profit: $P'(x) = 7 - 3x = 0 \Rightarrow x = \frac{7}{3} \approx 2.33$

Maximum profit: $P\left(\frac{7}{3}\right) = 7\left(\frac{7}{3}\right) - \frac{3}{2}\left(\frac{49}{9}\right) - 100$

$$= \frac{49}{3} - \frac{147}{18} - 100 = \frac{294}{18} - \frac{147}{18} - \frac{1800}{18} = -\frac{1653}{18} \approx -91.83$$

Actually negative profit at optimum means they should shut down!

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8. Average Value of a Function

8.1 Formula

Average value of $f$ on $[a, b]$:

$$f_{\text{avg}} = \frac{1}{b-a} \int_{a}^{b} f(x) dx$$

8.2 Mean Value Theorem for Integrals

If $f$ is continuous on $[a, b]$, then there exists $c$ in $(a, b)$ such that:

$$f(c) = f_{\text{avg}} = \frac{1}{b-a} \int_{a}^{b} f(x) dx$$

8.3 Example

Find average value of $f(x) = \sin x$ on $[0, \pi]$

Solution:

$$f_{\text{avg}} = \frac{1}{\pi - 0} \int_{0}^{\pi} \sin x dx = \frac{1}{\pi} [-\cos x]_{0}^{\pi}$$

$$= \frac{1}{\pi} (-\cos \pi + \cos 0) = \frac{1}{\pi} (1 + 1) = \frac{2}{\pi}$$

Find $c$ such that $\sin c = \frac{2}{\pi}$:

$c = \sin^{-1}\left(\frac{2}{\pi}\right) \approx 0.69$ radians

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9. Center of Mass (Centroid)

9.1 Centroid of a Region

For region bounded by $y = f(x)$, $y = 0$, $x = a$, $x = b$:

$$\bar{x} = \frac{1}{A} \int_{a}^{b} x f(x) dx$$

$$\bar{y} = \frac{1}{2A} \int_{a}^{b} [f(x)]^2 dx$$

where $A = \int_{a}^{b} f(x) dx$ = area

9.2 Example

Find centroid of region under $y = 4 - x^2$ from $x = -2$ to $x = 2$

Solution:

First find area:

$$A = \int_{-2}^{2} (4 - x^2) dx = \left[4x - \frac{x^3}{3}\right]_{-2}^{2}$$

$$= \left(8 - \frac{8}{3}\right) - \left(-8 + \frac{8}{3}\right) = \left(\frac{16}{3}\right) - \left(-\frac{16}{3}\right) = \frac{32}{3}$$

Now $\bar{x}$: By symmetry, $\bar{x} = 0$

Now $\bar{y}$:

$$\bar{y} = \frac{1}{2A} \int_{-2}^{2} (4 - x^2)^2 dx = \frac{3}{64} \int_{-2}^{2} (16 - 8x^2 + x^4) dx$$

Since even function: $= \frac{3}{32} \int_{0}^{2} (16 - 8x^2 + x^4) dx$

$$= \frac{3}{32} \left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]_{0}^{2}$$

$$= \frac{3}{32} \left(32 - \frac{64}{3} + \frac{32}{5}\right) = \frac{3}{32} \left(\frac{480 - 320 + 96}{15}\right)$$

$$= \frac{3}{32} \cdot \frac{256}{15} = \frac{768}{480} = \frac{8}{5} = 1.6$$

Centroid: $(0, 1.6)$

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10. Probability Density Functions

10.1 Properties of PDF

A probability density function $f(x)$ must satisfy:

1. $f(x) \geq 0$ for all $x$

2. $\int_{-\infty}^{\infty} f(x) dx = 1$

10.2 Probability Calculations

Probability that $X$ is between $a$ and $b$:

$$P(a \leq X \leq b) = \int_{a}^{b} f(x) dx$$

10.3 Mean (Expected Value)

$$\mu = E[X] = \int_{-\infty}^{\infty} x f(x) dx$$

10.4 Example: Uniform Distribution

$f(x) = \frac{1}{b-a}$ for $a \leq x \leq b$, 0 otherwise

Verify: $\int_{-\infty}^{\infty} f(x) dx = \int_{a}^{b} \frac{1}{b-a} dx = 1$

Mean: $\mu = \int_{a}^{b} x \cdot \frac{1}{b-a} dx = \frac{1}{b-a} \left[\frac{x^2}{2}\right]_{a}^{b} = \frac{b^2 - a^2}{2(b-a)} = \frac{a+b}{2}$

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11. Hydrostatic Force

11.1 Formula

Force on vertical plate submerged in fluid:

$$F = \int_{a}^{b} \rho g \cdot \text{depth}(x) \cdot \text{width}(x) dx$$

where:

• $\rho$ = density of fluid

• $g$ = gravitational acceleration

• depth(x) = depth at position x

• width(x) = width of plate at position x

11.2 Example

Find hydrostatic force on rectangular plate 2m wide × 3m high, submerged vertically with top 1m below water surface (ρg = 9800 N/m³)

Solution: Place origin at water surface, x positive downward.

Plate extends from $x = 1$ to $x = 4$ (depth 1m to 4m)

Width = 2m (constant)

$$F = \int_{1}^{4} 9800 \cdot x \cdot 2 dx = 19600 \int_{1}^{4} x dx$$

$$= 19600 \left[\frac{x^2}{2}\right]_{1}^{4} = 9800 (16 - 1) = 9800 \cdot 15 = 147000 \text{ N}$$

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12. Work and Energy

12.1 Work Done by Variable Force

Work = $\int_{a}^{b} F(x) dx$ where $F(x)$ varies with position

12.2 Example: Spring

Work to stretch spring from natural length to extension $L$:

$F(x) = kx$ (Hooke's Law)

$$W = \int_{0}^{L} kx dx = \frac{1}{2}kL^2$$

12.3 Example: Pumping Liquid

Work to pump liquid from tank:

1. Slice liquid into thin horizontal slabs

2. Find work to lift each slab to top

3. Integrate

Example: Cylindrical tank radius 2m, height 5m, full of water (ρg = 9800 N/m³). Find work to pump all water to top.

Solution: Place origin at bottom of tank. Slice at height $x$, thickness $dx$.

Volume of slice = π(2)² dx = 4π dx m³

Weight = 9800 × 4π dx = 39200π dx N

Distance to lift = (5 - x) m

Work for slice = 39200π(5 - x) dx

Total work:

$$W = \int_{0}^{5} 39200\pi (5 - x) dx = 39200\pi \left[5x - \frac{x^2}{2}\right]_{0}^{5}$$

$$= 39200\pi \left(25 - \frac{25}{2}\right) = 39200\pi \cdot \frac{25}{2} = 490000\pi \text{ J}$$

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13. Important Formulas Summary

13.1 Area Formulas

1. Area under curve: $A = \int_{a}^{b} f(x) dx$ (net area)

2. Area between curves: $A = \int_{a}^{b} [f(x) - g(x)] dx$

3. Total area: $A = \int_{a}^{b} |f(x)| dx$

13.2 Volume Formulas

1. Disk method: $V = \pi \int_{a}^{b} [f(x)]^2 dx$

2. Washer method: $V = \pi \int_{a}^{b} ([f(x)]^2 - [g(x)]^2) dx$

3. Shell method: $V = 2\pi \int_{a}^{b} x f(x) dx$

13.3 Physics Formulas

1. Displacement: $\int v(t) dt$

2. Distance: $\int |v(t)| dt$

3. Work: $\int F(x) dx$

4. Average value: $\frac{1}{b-a} \int_{a}^{b} f(x) dx$

13.4 Economics Formulas

1. Total cost: $C(x) = \int C'(x) dx + C_0$

2. Total revenue: $R(x) = \int R'(x) dx$

3. Consumer surplus: $\int_{0}^{q^*} [D(q) - p^*] dq$

4. Producer surplus: $\int_{0}^{q^*} [p^* - S(q)] dq$

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14. Solved Examples

Example 1: Complex Area Problem

Find area enclosed by $y = x^2$ and $y = 2x - x^2$

Solution: Intersection: $x^2 = 2x - x^2 \Rightarrow 2x^2 - 2x = 0 \Rightarrow 2x(x-1) = 0 \Rightarrow x=0,1$

Between 0 and 1: $2x - x^2 \geq x^2$

$$A = \int_{0}^{1} [(2x - x^2) - x^2] dx = \int_{0}^{1} (2x - 2x^2) dx$$

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

Example 2: Volume with Hole

Find volume when region between $y = x^2$ and $y = x$ is rotated about y-axis

Solution: Intersection: $x^2 = x \Rightarrow x(x-1)=0 \Rightarrow x=0,1$

Points: (0,0) and (1,1)

Using shell method:

Radius = $x$, height = $x - x^2$

$$V = 2\pi \int_{0}^{1} x(x - x^2) dx = 2\pi \int_{0}^{1} (x^2 - x^3) dx$$

$$= 2\pi \left[\frac{x^3}{3} - \frac{x^4}{4}\right]_{0}^{1} = 2\pi \left(\frac{1}{3} - \frac{1}{4}\right) = 2\pi \cdot \frac{1}{12} = \frac{\pi}{6}$$

Example 3: Work Problem

Chain 10m long, mass 2kg/m, hangs from building. Find work to pull it up.

Solution: Let $x$ = distance from top. Slice at position $x$, length $dx$.

Mass of slice = 2 dx kg Weight = 2g dx = 19.6 dx N (taking g=9.8)

Distance to lift = $x$ m

Work for slice = 19.6x dx

Total work:

$$W = \int_{0}^{10} 19.6x dx = 19.6 \left[\frac{x^2}{2}\right]_{0}^{10} = 9.8 \times 100 = 980 \text{ J}$$

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15. Common Mistakes and Exam Tips

15.1 Common Mistakes

1. Forgetting absolute value when computing total area

2. Mixing up radius and height in shell/disk methods

3. Using wrong limits of integration

4. Forgetting units in applied problems

5. Not checking which function is above/between curves

15.2 Problem-Solving Strategy

1. Draw diagram: Always sketch the situation

2. Set up coordinate system: Choose origin and axes wisely

3. Slice: Think about slicing the region/object

4. Write element: Express area/volume/work of slice

5. Integrate: Set up and evaluate definite integral

6. Check: Does answer make sense? Units correct?

15.3 Quick Checks

1. Area: Should be positive

2. Volume: Should be positive

3. Work/Energy: Check sign (work done ON system vs BY system)

4. Symmetry: Use to simplify problems

5. Dimensional analysis: Check units make sense

This comprehensive theory covers all applications of antiderivatives with detailed explanations and examples, providing complete preparation for the entrance examination.