Quiz 2

Definite Integrals and the Fundamental Theorem of Calculus

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# Definite Integrals and the Fundamental Theorem of Calculus ## 🎯 Learning Objectives By the end of this topic, you will be able to: 1. **Define** the definite integral as the limit of Riemann sums 2.

Definite Integrals and the Fundamental Theorem of Calculus

🎯 Learning Objectives

By the end of this topic, you will be able to:
  1. Define the definite integral as the limit of Riemann sums
  2. State the Fundamental Theorem of Calculus (Parts 1 and 2)
  3. Evaluate definite integrals using the FTC
  4. Compute the area between curves
  5. Apply properties of definite integrals

📋 Prerequisites

📖 Core Content

16.1 Intuition: Accumulation of Area

The definite integral ∫abf(x) dx\int_a^b f(x)\,dx computes the signed area between f(x)f(x) and the xx-axis from x=ax = a to x=bx = b. Area above the axis is positive; area below is negative. Imagine tracking the total distance traveled by a car whose speed varies: speed v(t)v(t) at time tt, total distance = ∫abv(t) dt\int_a^b v(t)\,dt, the area under the speed curve.
💡 Why this matters: Definite integrals compute accumulated quantities: total distance from velocity, total charge from current, total probability from density, expected value in statistics.

16.2 Riemann Sums

Divide [a,b][a, b] into nn subintervals of width Δx=(b−a)/n\Delta x = (b-a)/n. Choose a sample point xi∗x_i^* in each subinterval. Then:
∫abf(x) dx=lim⁡n→∞∑i=1nf(xi∗) Δx\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*)\, \Delta x

16.3 Fundamental Theorem of Calculus (FTC)

Part 1 (FTC1): If F′(x)=f(x)F'(x) = f(x), then ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a). Part 2 (FTC2): If g(x)=∫axf(t) dtg(x) = \int_a^x f(t)\,dt, then g′(x)=f(x)g'(x) = f(x). FTC1 connects anti-derivatives to areas. FTC2 says differentiation undoes integration (and vice versa).

16.4 Properties of Definite Integrals

PropertyFormula
Reverse limits∫abf(x) dx=−∫baf(x) dx\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx
Zero width∫aaf(x) dx=0\int_a^a f(x)\,dx = 0
Constant multiple∫abcf(x) dx=c∫abf(x) dx\int_a^b c f(x)\,dx = c\int_a^b f(x)\,dx
Sum∫ab[f(x)+g(x)] dx=∫abf(x) dx+∫abg(x) dx\int_a^b [f(x)+g(x)]\,dx = \int_a^b f(x)\,dx + \int_a^b g(x)\,dx
Additivity∫acf(x) dx=∫abf(x) dx+∫bcf(x) dx\int_a^c f(x)\,dx = \int_a^b f(x)\,dx + \int_b^c f(x)\,dx
ComparisonIf f(x)≤g(x)f(x) \leq g(x) on [a,b][a,b] , then ∫abf(x) dx≤∫abg(x) dx\int_a^b f(x)\,dx \leq \int_a^b g(x)\,dx

16.5 Area Between Curves

Area between y=f(x)y = f(x) (top) and y=g(x)y = g(x) (bottom) from x=ax = a to x=bx = b:
Area=∫ab[f(x)−g(x)] dx\text{Area} = \int_a^b [f(x) - g(x)]\,dx

16.6 Worked Examples

Example 1.1 (FTC): Evaluate ∫02x2 dx\int_0^2 x^2\,dx.
∫02x2 dx=[x33]02=83−0=83\int_0^2 x^2\,dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3} - 0 = \frac{8}{3} 83\boxed{\frac{8}{3}}
Example 1.2: Evaluate ∫1e1x dx\int_1^e \frac{1}{x}\,dx.
∫1e1x dx=[ln⁡x]1e=ln⁡e−ln⁡1=1−0=1\int_1^e \frac{1}{x}\,dx = [\ln x]_1^e = \ln e - \ln 1 = 1 - 0 = 1 1\boxed{1}
Example 1.3 (Area between curves): Find the area between y=xy = x and y=x2y = x^2 from x=0x = 0 to x=1x = 1. Area=∫01(x−x2) dx=[x22−x33]01=12−13=16\text{Area} = \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}
16\boxed{\frac{1}{6}}
Example 1.4: Find ∫02∣x−1∣ dx\int_0^2 |x-1|\,dx.
∣x−1∣={1−x0≤x≤1x−11≤x≤2|x-1| = \begin{cases} 1-x & 0 \leq x \leq 1 \\ x-1 & 1 \leq x \leq 2 \end{cases} ∫02∣x−1∣ dx=∫01(1−x) dx+∫12(x−1) dx=[x−x22]01+[x22−x]12=12+12=1\int_0^2 |x-1|\,dx = \int_0^1 (1-x)\,dx + \int_1^2 (x-1)\,dx = \left[x - \frac{x^2}{2}\right]_0^1 + \left[\frac{x^2}{2} - x\right]_1^2 = \frac{1}{2} + \frac{1}{2} = 1 1\boxed{1}

📐 Key Formulas — Summary Table

ConceptFormula
FTC∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a) where F′=fF' = f
Area between curves∫ab[f(x)−g(x)] dx\int_a^b [f(x) - g(x)]\,dx (top minus bottom)
Average value1b−a∫abf(x) dx\frac{1}{b-a}\int_a^b f(x)\,dx
Additivity∫ab+∫bc=∫ac\int_a^b + \int_b^c = \int_a^c
PiecewiseSplit integral at piece boundaries

⚠️ Common Pitfalls

Pitfall 1: Forgetting to Subtract F(a)F(a)

∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a), NOT F(b)F(b) alone.

Pitfall 2: Confusing Definite and Indefinite

Definite integrals have limits and produce a NUMBER. Indefinite integrals have no limits and produce a FAMILY of functions.

Pitfall 3: Area For Absolute Value

For ∫∣f(x)∣ dx\int |f(x)|\,dx, split at points where f(x)=0f(x) = 0 and integrate each piece separately.

📝 Practice Questions

Q1: Evaluate ∫132x dx\int_1^3 2x\,dx.
[x2]13=9−1=8[x^2]_1^3 = 9 - 1 = 8
8\boxed{8} Q2: Evaluate ∫0πsin⁡x dx\int_0^{\pi} \sin x\,dx.
[−cos⁡x]0π=−(−1)−(−1)=1+1=2[-\cos x]_0^{\pi} = -(-1) - (-1) = 1 + 1 = 2
2\boxed{2} Q3: Find area under y=exy = e^x from x=0x = 0 to x=1x = 1.
∫01ex dx=[ex]01=e−1\int_0^1 e^x\,dx = [e^x]_0^1 = e - 1
e−1\boxed{e - 1} Q4: Find the area between y=x2y = x^2 and y=x+2y = x + 2.
Intersection: x2=x+2→x2−x−2=0→(x−2)(x+1)=0→x=−1,2x^2 = x + 2 \to x^2 - x - 2 = 0 \to (x-2)(x+1) = 0 \to x = -1, 2 ∫−12[(x+2)−x2] dx=[x22+2x−x33]−12=92\int_{-1}^2 [(x+2) - x^2]\,dx = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^2 = \frac{9}{2}
4.5\boxed{4.5} Q5: Evaluate ∫02(x3−2x) dx\int_0^2 (x^3 - 2x)\,dx.
[x44−x2]02=(4−4)−0=0\left[\frac{x^4}{4} - x^2\right]_0^2 = (4 - 4) - 0 = 0
0\boxed{0} Q6: Find ∫−11∣x∣ dx\int_{-1}^1 |x|\,dx.
2∫01x dx=2[x22]01=12\int_0^1 x\,dx = 2\left[\frac{x^2}{2}\right]_0^1 = 1
1\boxed{1} Q7: Evaluate ∫141x dx\int_1^4 \frac{1}{\sqrt{x}}\,dx.
∫14x−1/2 dx=[2x]14=2(2)−2(1)=2\int_1^4 x^{-1/2}\,dx = [2\sqrt{x}]_1^4 = 2(2) - 2(1) = 2
2\boxed{2} Q8: Find ∫0111+x2 dx\int_0^1 \frac{1}{1+x^2}\,dx.
[arctan⁡x]01=π4−0=π4[\arctan x]_0^1 = \frac{\pi}{4} - 0 = \frac{\pi}{4}
π4\boxed{\frac{\pi}{4}} Q9: Average value of f(x)=x2f(x) = x^2 on [0,3][0, 3].
13∫03x2 dx=13[x33]03=13⋅9=3\frac{1}{3}\int_0^3 x^2\,dx = \frac{1}{3}\left[\frac{x^3}{3}\right]_0^3 = \frac{1}{3} \cdot 9 = 3
3\boxed{3} Q10: Find ∫01e2x dx\int_0^1 e^{2x}\,dx.
[e2x2]01=e22−12=e2−12\left[\frac{e^{2x}}{2}\right]_0^1 = \frac{e^2}{2} - \frac{1}{2} = \frac{e^2 - 1}{2}
e2−12\boxed{\frac{e^2 - 1}{2}}

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