# 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: $$\int\_{a}^{b} f(x) dx = F(b) - F(a)$$

***

### **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

***

### **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}$$

***

### **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}
$$

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

***

### **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)
$$

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

***

### **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}
$$

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

2. 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 $$x\_2$$:

$$
W = \int\_{x\_1}^{x\_2} kx dx = \frac{k}{2}(x\_2^2 - x\_1^2)
$$

***

### **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)$$

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

4. 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!

***

### **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

***

### **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: $$(0, 1.6)$$

***

### **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}$$

***

### **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}
$$

***

### **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}
$$

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

***

### **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$$

***

### **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}
$$

***

### **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.

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