1. LECTURE 7
Antidifferentiation

2. Objectives
Antiderivative
Indefinite Integrals
Integration Rules

3. Introduction Two fundamental problems in Calculus
Slope of a curve at a point – rate of change – derivative.
Area of a region under a curve – integral of a function
Their connection: Fundamental Theorem of Calculus.

4. Introduction
Example: During the early 1970s, the annual worldwide rate of oil consumption grew exponentially with a growth constant of about At the beginning of 1970, the rate was about billion barrels per year. Let R(t) denote the rate of oil consumption at time t. Then a reasonable model for R(t) is given by: R(t) = e0.07t. Find the amount of oil that would have consumed from to 1980.
Let T(t) be the amount of oil consumed from time 0 (1970) to time t ( t)  T’(t) = R(t)  T(t): an anti-derivative of R(t).
Problem of anti-differentiation

5. Antiderivative F(x) = x2  f(x) = F’(x) = 2x
f(x) is the derivative of F(x)
F(x) is an anti-derivative of f(x)

6. Antiderivative F1(x) = x2  F1’(x) = 2x F2(x) = x2 +3  F2’(x) = 2x
Conclusion:

7. Antiderivative
Theorem I: Suppose that f(x) is a continuous function on an interval I.
If F1(x) and F2(x) are two anti-derivatives of f(x) then F1(x) and F2(x) differ by a constant on I.
In other words, there is a constant C such that
F2(x) = F1(x) + C.
Theorem II: If F’(x) = 0 for all x in an interval I, then there is a constant C such that F(x) = C for all x in I

8. Indefinite Integral If F(x) is an antiderivative of f(x) then
∫f(x)dx = F(x) +C
∫: integral sign
∫f(x)dx: indefinite integral

9. Integration Rules Rule 1: ∫kdx = kx +C
Rule 2: ∫xndx = xn+1/(n+1) +C (n≠-1)
∫x3dx =
∫x-3dx =
∫x0.5dx =
∫dx =

10. Integration Rules Rule 3: ∫kf(x)dx = k∫f(x)dx
Rule 4: ∫[f(x)±g(x)]dx = ∫f(x)dx ± ∫g(x)dx
∫[5x3-3x5]dx =
∫(2/3x5+4x0.5)dx =

11. Integration Rules Rule 5: ∫x-1dx = lnӏxІ+C Rule 6: ∫axdx = ax/lna+C
∫exdx = ex+C
∫2exdx=
∫[(2x+1)/x]dx=
∫2xexdx=

12. Integration Rules Rule 7: ∫cosxdx = sinx+C Rule 8: ∫sinxdx = -cosx+C
Rule 9: ∫ 1 𝑐𝑜𝑠 2 𝑥 dx = tanx+C
Rule 10: ∫ 1 𝑠𝑖𝑛 2 𝑥 dx = -cotx+C

13. Integration by Substitution
Let u=u(x)
∫f(x)dx= ∫ g[u(x)]u’(x) dx = ∫ g(u)du
Example:
I= ∫[(2x2+5x+1)2016(4x+5)]dx

14. Integration by Substitution
Let x=x(t)
∫f(x)dx= ∫f(x(t))x’(t)dt= ∫ g(t)dt
Example:
I= ∫dx/(1+x2)

15. Integration by Parts ∫f(x)g’(x)dx= f(x)g(x)- ∫f’(x)g(x)dx
∫(2x+1) exdx=
∫(2x+1) lnxdx=

16. Example 1
Oil problem

17. Example 2
A rocket is fired vertically into the air. Its velocity at t seconds after lift-off is v(t) = 6t m/s. Before launch, the top of the rocket is 8 meters above the launch pad. Find the height of the rocket (measured from the top of the rocket to the launch pad) at time t.

18. Example 3
A company’s marginal cost function is 0.015x2 – 2x dollars, where x denotes the number of units produced in one day. The company has fixed costs of $1000 per day.
Find the cost of producing x units per day.
Suppose current level of production is x = 30. Determine the amount costs will rise if the production level is raised to x = 60 units.

19. Solution