Unit 6 · Topic 6.7 Beta

The Fundamental Theorem and Definite Integrals

To evaluate ∫ from a to b of f(x) dx , find an antiderivative F and compute F(b) − F(a) .

Practice 1: Implementing Mathematical Processes

Question set for this topic

Part 1 · Hook

Why this matters

Finding an area by stacking millions of thin rectangles sounds hopeless by hand. Yet ∫13(3x2−2)dx\int_1^3 (3x^2 - 2)\,dx takes four lines: find a function whose derivative is 3x2−23x^2 - 2, plug in 3 and 1, subtract. That shortcut, the second half of the Fundamental Theorem, is why calculus can compute areas, distances and totals at all.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If g(x)g(x)=∫axf(t)dt= \int_a^x f(t)\,dt and ff is continuous, which is true?

  1. gg is an antiderivative of ff
  2. ff is an antiderivative of gg
  3. g=f′g = f'
  4. g(a)=f(a)g(a) = f(a)
Show the answer

g′(x)=f(x)g'(x) = f(x), so gg is an antiderivative of ff.

  • Correct: gg is an antiderivative of ff:
  • ff is an antiderivative of gg:
  • g=f′g = f':
  • g(a)=f(a)g(a) = f(a):

2. If ∫02f(x)dx\int_0^2 f(x)\,dx=3= 3, what is ∫20f(x)dx\int_2^0 f(x)\,dx?

  1. −3-3
  2. 3
  3. 0
  4. 6
Show the answer

Reversing the limits changes the sign.

  • Correct: −3-3:
  • 3:
  • 0:
  • 6:

3. What is the derivative of arctan⁡x\arctan x?

  1. 11+x2\frac{1}{1 + x^2}
  2. sec2⁡x\sec^2 x
  3. 11−x2\frac{1}{\sqrt{1 - x^2}}
  4. −11+x2\frac{-1}{1 + x^2}
Show the answer

The inverse tangent derivative.

  • Correct: 11+x2\frac{1}{1 + x^2}:
  • sec2⁡x\sec^2 x:
  • 11−x2\frac{1}{\sqrt{1 - x^2}}:
  • −11+x2\frac{-1}{1 + x^2}:

Part 4 · See it

See it first

Top: the graph of f prime of x = 1 - x/4 on [0, 6], with a triangle of area 2 above the axis from 0 to 4 and a triangle of area 0.5 below from 4 to 6. Bottom: f(x) = 1 + x - x squared over 8 rises from (0, 1) to a maximum at x = 4 and falls to (6, 2.5); a bracket shows f(6) - f(0) = 2 - 0.5 = 1.5.
The area under f prime from 0 to 6 (2 above, 0.5 below) equals f(6) - f(0) = 1.5. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. g(x)g(x)=∫axf(t)dt= \int_a^x f(t)\,dt and any FF are both antiderivatives of ffthey differ by a constant: g(x)=F(x)−F(a)g(x) = F(x) - F(a), since g(a)=0g(a) = 0
  2. Put x=bx = b∫abf(x)dx\int_a^b f(x)\,dx=F(b)−F(a)= F(b) - F(a), the Fundamental Theorem, part 2
  3. Apply it to a rate f′f'∫abf′(x)dx\int_a^b f'(x)\,dx=f(b)−f(a)= f(b) - f(a): the integral of a rate is the net change
  4. You know f(a)f(a) and the graph of f′f'f(b)f(b)=f(a)= f(a)+∫abf′(x)dx+ \int_a^b f'(x)\,dx, with signed areas from the graph

Part 6 · Key ideas

Key ideas

  • FTC, part 2: if F′=fF' = f and ff is continuous on [a,b][a, b], then ∫abf(x)dx\int_a^b f(x)\,dx=F(b)−F(a)= F(b) - F(a)=F(x)|ab= F(x)\Big|_a^b.
  • Net change: ∫abf′(x)dx\int_a^b f'(x)\,dx=f(b)−f(a)= f(b) - f(a), so f(b)f(b)=f(a)= f(a)+∫abf′(x)dx+ \int_a^b f'(x)\,dx.
  • Going backward from a known value: f(0)f(0)=f(2)−∫02f′(x)dx= f(2) - \int_0^2 f'(x)\,dx.
  • With no elementary antiderivative, a calculator evaluates the integral; write the integral as the setup.

Part 7 · Misconception

A common mistake

The wrong idea: f(5)f(5)=∫25f′(x)dx= \int_2^5 f'(x)\,dx: the integral of the derivative gives the function.

What actually happens: The integral of f′f' gives the change in ff. Add the known starting value: f(5)f(5)=f(2)= f(2)+∫25f′(x)dx+ \int_2^5 f'(x)\,dx.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of f′f', the derivative of ff

The graph of f′f', the derivative of a function ff, is shown on [0,8][0, 8]; it is made of line segments. f(2)=5f(2) = 5.
12345678−2−112xyf'

1. Find f(5)f(5).

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

f(5)f(5)=f(2)= f(2)+∫25f′(x)dx+ \int_2^5 f'(x)\,dx=5= 5+(2+1−1)+ (2 + 1 - 1)=7= 7. (Rectangle 2 to 3: 2; triangle 3 to 4: 1 above; triangle 4 to 5: 1 below.)

  • Answer: 7

2. Working backward from f(2)=5f(2) = 5, find f(0)f(0).

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

f(0)f(0)=f(2)= f(2)+∫20f′(x)dx+ \int_2^0 f'(x)\,dx=5−∫02f′(x)dx= 5 - \int_0^2 f'(x)\,dx=5−4= 5 - 4=1= 1.

  • Answer: 1

3. Find f(8)f(8).

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

∫28f′(x)dx\int_2^8 f'(x)\,dx=2= 2+1−1−6+ 1 - 1 - 6=−4= -4, so f(8)=5+(−4)=1f(8) = 5 + (-4) = 1.

  • Answer: 1

4. What is the absolute maximum value of ff on [0,8][0, 8]?

  1. 7, at x = 5
  2. 8, at x = 4
  3. 5, at x = 2
  4. 2, at x = 3
Show the answer

f′=0f' = 0 and changes from positive to negative at x=4x = 4. Candidates: f(0)=1f(0) = 1, f(4)f(4)=f(2)= f(2)+∫24f′(x)dx+ \int_2^4 f'(x)\,dx=5= 5+3+ 3=8= 8, f(8)=1f(8) = 1. The absolute maximum is 8, at x=4x = 4.

  • 7, at x = 5: f(5)=7f(5) = 7, but f′<0f' < 0 on (4,5)(4, 5): ff was higher at 4.
  • Correct: 8, at x = 4: Correct: candidates f(0)=1f(0) = 1, f(4)=5+2+1=8f(4) = 5 + 2 + 1 = 8, f(8)=1f(8) = 1.
  • 5, at x = 2: f′>0f' > 0 just after 2, so ff keeps growing until f′f' changes sign at x=4x = 4.
  • 2, at x = 3: 2 is the largest value of f′f', the rate, not of ff.

5. Find ∫13(3x2−2)dx\int_1^3 (3x^2 - 2)\,dx.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

An antiderivative is F(x)=x3−2xF(x) = x^3 - 2x. ∫13(3x2−2)dx\int_1^3 (3x^2 - 2)\,dx=F(3)−F(1)= F(3) - F(1)=21−(−1)= 21 - (-1)=22= 22.

  • Answer: 22

6. hh is differentiable, h(1)=4h(1) = 4 and ∫16h′(x)dx\int_1^6 h'(x)\,dx=−3= -3. What is h(6)h(6)?

  1. −3-3
  2. 7
  3. −7-7
  4. 1
Show the answer

h(6)h(6)=h(1)= h(1)+∫16h′(x)dx+ \int_1^6 h'(x)\,dx=4= 4+(−3)+ (-3)=1= 1.

  • −3-3: That is the change h(6)−h(1)h(6) - h(1). Add it to h(1)=4h(1) = 4.
  • 7: The change is −3-3: the value goes down by 3, not up.
  • −7-7: h(6)h(6)=h(1)= h(1)+∫16h′(x)dx+ \int_1^6 h'(x)\,dx, not h(1)h(1) subtracted from the change.
  • Correct: 1: Correct: 4 + (-3).

Part 9 · Summary

Summary

To evaluate ∫abf(x)dx\int_a^b f(x)\,dx, find an antiderivative FF and compute F(b)−F(a)F(b) - F(a). Read in context, the integral of a rate of change is the net change: f(b)f(b)=f(a)= f(a)+∫abf′(x)dx+ \int_a^b f'(x)\,dx. Use signed areas from a graph of f′f', an antiderivative from a formula, or a calculator when no formula exists.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections