Definite Integrals and the Fundamental Theorem of Calculus

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# Definite Integrals and the Fundamental Theorem of Calculus ## ๐ŸŽฏ 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

๐ŸŽฏ 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

๐Ÿ“‹ Prerequisites


๐Ÿ“– Core Content

16.1 Intuition: Accumulation of Area

The definite integral โˆซabf(x)โ€‰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 = โˆซabv(t)โ€‰dt\int_a^b v(t)\,dt, the area under the speed curve.
๐Ÿ’ก 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 ฮ”x=(bโˆ’a)/n\Delta x = (b-a)/n. Choose a sample point xiโˆ—x_i^* in each subinterval. Then:
โˆซabf(x)โ€‰dx=limโกnโ†’โˆžโˆ‘i=1nf(xiโˆ—)โ€‰ฮ”x\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 Fโ€ฒ(x)=f(x)F'(x) = f(x), then โˆซabf(x)โ€‰dx=F(b)โˆ’F(a)\int_a^b f(x)\,dx = F(b) - F(a). Part 2 (FTC2): If g(x)=โˆซaxf(t)โ€‰dtg(x) = \int_a^x f(t)\,dt, then gโ€ฒ(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 limitsโˆซabf(x)โ€‰dx=โˆ’โˆซbaf(x)โ€‰dx\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx
Zero widthโˆซaaf(x)โ€‰dx=0\int_a^a f(x)\,dx = 0
Constant multipleโˆซabcf(x)โ€‰dx=cโˆซabf(x)โ€‰dx\int_a^b c f(x)\,dx = c\int_a^b f(x)\,dx
Sumโˆซab[f(x)+g(x)]โ€‰dx=โˆซabf(x)โ€‰dx+โˆซabg(x)โ€‰dx\int_a^b [f(x)+g(x)]\,dx = \int_a^b f(x)\,dx + \int_a^b g(x)\,dx
Additivityโˆซacf(x)โ€‰dx=โˆซabf(x)โ€‰dx+โˆซbcf(x)โ€‰dx\int_a^c f(x)\,dx = \int_a^b f(x)\,dx + \int_b^c f(x)\,dx
ComparisonIf f(x)โ‰คg(x)f(x) \leq g(x) on [a,b][a,b] , then โˆซabf(x)โ€‰dxโ‰คโˆซabg(x)โ€‰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=โˆซab[f(x)โˆ’g(x)]โ€‰dx\text{Area} = \int_a^b [f(x) - g(x)]\,dx

16.6 Worked Examples

Example 1.1 (FTC): Evaluate โˆซ02x2โ€‰dx\int_0^2 x^2\,dx.
โˆซ02x2โ€‰dx=[x33]02=83โˆ’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 โˆซ1e1xโ€‰dx\int_1^e \frac{1}{x}\,dx.
โˆซ1e1xโ€‰dx=[lnโกx]1e=lnโกeโˆ’lnโก1=1โˆ’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=โˆซ01(xโˆ’x2)โ€‰dx=[x22โˆ’x33]01=12โˆ’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 โˆซ02โˆฃxโˆ’1โˆฃโ€‰dx\int_0^2 |x-1|\,dx.
โˆฃxโˆ’1โˆฃ={1โˆ’x0โ‰คxโ‰ค1xโˆ’11โ‰คxโ‰ค2|x-1| = \begin{cases} 1-x & 0 \leq x \leq 1 \\ x-1 & 1 \leq x \leq 2 \end{cases} โˆซ02โˆฃxโˆ’1โˆฃโ€‰dx=โˆซ01(1โˆ’x)โ€‰dx+โˆซ12(xโˆ’1)โ€‰dx=[xโˆ’x22]01+[x22โˆ’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}

๐Ÿ“ Key Formulas โ€” Summary Table

ConceptFormula
FTCโˆซabf(x)โ€‰dx=F(b)โˆ’F(a)\int_a^b f(x)\,dx = F(b) - F(a) where Fโ€ฒ=fF' = f
Area between curvesโˆซab[f(x)โˆ’g(x)]โ€‰dx\int_a^b [f(x) - g(x)]\,dx (top minus bottom)
Average value1bโˆ’aโˆซabf(x)โ€‰dx\frac{1}{b-a}\int_a^b f(x)\,dx
Additivityโˆซab+โˆซbc=โˆซac\int_a^b + \int_b^c = \int_a^c
PiecewiseSplit integral at piece boundaries

โš ๏ธ Common Pitfalls

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

โˆซabf(x)โ€‰dx=F(b)โˆ’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 โˆซโˆฃf(x)โˆฃโ€‰dx\int |f(x)|\,dx, split at points where f(x)=0f(x) = 0 and integrate each piece separately.

๐Ÿ“ Practice Questions

Q1: Evaluate โˆซ132xโ€‰dx\int_1^3 2x\,dx.
[x2]13=9โˆ’1=8[x^2]_1^3 = 9 - 1 = 8
8\boxed{8} Q2: Evaluate โˆซ0ฯ€sinโกxโ€‰dx\int_0^{\pi} \sin x\,dx.
[โˆ’cosโกx]0ฯ€=โˆ’(โˆ’1)โˆ’(โˆ’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.
โˆซ01exโ€‰dx=[ex]01=eโˆ’1\int_0^1 e^x\,dx = [e^x]_0^1 = e - 1
eโˆ’1\boxed{e - 1} Q4: Find the area between y=x2y = x^2 and y=x+2y = x + 2.
Intersection: x2=x+2โ†’x2โˆ’xโˆ’2=0โ†’(xโˆ’2)(x+1)=0โ†’x=โˆ’1,2x^2 = x + 2 \to x^2 - x - 2 = 0 \to (x-2)(x+1) = 0 \to x = -1, 2 โˆซโˆ’12[(x+2)โˆ’x2]โ€‰dx=[x22+2xโˆ’x33]โˆ’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 โˆซ02(x3โˆ’2x)โ€‰dx\int_0^2 (x^3 - 2x)\,dx.
[x44โˆ’x2]02=(4โˆ’4)โˆ’0=0\left[\frac{x^4}{4} - x^2\right]_0^2 = (4 - 4) - 0 = 0
0\boxed{0} Q6: Find โˆซโˆ’11โˆฃxโˆฃโ€‰dx\int_{-1}^1 |x|\,dx.
2โˆซ01xโ€‰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 โˆซ141xโ€‰dx\int_1^4 \frac{1}{\sqrt{x}}\,dx.
โˆซ14xโˆ’1/2โ€‰dx=[2x]14=2(2)โˆ’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 โˆซ0111+x2โ€‰dx\int_0^1 \frac{1}{1+x^2}\,dx.
[arctanโกx]01=ฯ€4โˆ’0=ฯ€4[\arctan x]_0^1 = \frac{\pi}{4} - 0 = \frac{\pi}{4}
ฯ€4\boxed{\frac{\pi}{4}} Q9: Average value of f(x)=x2f(x) = x^2 on [0,3][0, 3].
13โˆซ03x2โ€‰dx=13[x33]03=13โ‹…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 โˆซ01e2xโ€‰dx\int_0^1 e^{2x}\,dx.
[e2x2]01=e22โˆ’12=e2โˆ’12\left[\frac{e^{2x}}{2}\right]_0^1 = \frac{e^2}{2} - \frac{1}{2} = \frac{e^2 - 1}{2}
e2โˆ’12\boxed{\frac{e^2 - 1}{2}}

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