1. Distance Traveled Area Under a curve Antiderivatives
Calculus
Review
Distance Traveled
Area Under a curve
Antiderivatives

2. Consider an object moving at a constant rate of 3 ft/sec.
Since rate . time = distance:
If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line.
time
velocity
After 4 seconds, the object has gone 12 feet.

3. If the velocity is not constant,
we might guess that the
distance traveled is still equal
to the area under the curve.
(The units work out.)
Example:
We could estimate the area under the curve by drawing rectangles touching at their left corners.
This is called the Left-hand Rectangular Approximation Method (LRAM).
Approximate area:

4. We could also use a Right-hand Rectangular Approximation Method (RRAM).
Approximate area:

5. Another approach would be to use rectangles that touch at the midpoint
Another approach would be to use rectangles that touch at the midpoint. This is the Midpoint Rectangular Approximation Method (MRAM).
In this example there are four subintervals.
As the number of subintervals increases, so does the accuracy.
Approximate area:

6. The exact answer for this problem is .
With 8 subintervals:
Approximate area:
The exact answer for this
problem is
width of subinterval

7. Inscribed rectangles are all below the curve:
Circumscribed rectangles are all above the curve:

8. When we find the area under a curve by adding rectangles, the answer is called a Rieman sum.
The width of a rectangle is called a subinterval.
subinterval
Subintervals do not all have to be the same size.

9. variable of integration
upper limit of integration
Integration
Symbol
integrand
variable of integration
(dummy variable)
lower limit of integration
It is called a dummy variable because the answer does not depend on the variable chosen.

10. We have the notation for integration, but we still need to learn how to evaluate the integral.

11. Since rate . time = distance:
In section 5.1, we considered an object moving at a constant rate of 3 ft/sec.
Since rate . time = distance:
If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line.
time
velocity
After 4 seconds, the object has gone 12 feet.

12. If the velocity varies:
Distance:
(C=0 since s=0 at t=0)
After 4 seconds:
The distance is still equal to the area under the curve!
Notice that the area is a trapezoid.

13. Fill in the blanks:
Integrating velocity gives ____________________
Integrating the absolute value of velocity
gives ______________
c) New Position = _____________ + _______________
Displacement
Total Distance
Total Distance
Initial Disp.

14. Antiderivatives
Definition: A function F is called an antiderivative of f on an interval I if F’(x) = f(x) for all x in I.
Example: Let f(x)=x3. If F(x) =1/4 * x4 then F’(x) = f(x)
Theorem: If F is an antiderivative of f
on an interval I, then the most general
antiderivative of f on I is
F(x) + C
where C is an arbitrary constant.
F is an antiderivative of f

15. Table of particular antiderivatives
Function
Antiderivative
xr, r  -1
xr+1/(r+1)
1/x
ln |x|
sin x
- cos x ex
cos x ax
ax / ln a
sec2 x
tan x
tan-1 x
sin-1 x
These rules give particular antiderivatives of the listed functions. A general antiderivative can be obtained by adding a constant .

16. Integrate

17. Rewriting before integrating

18. 0 = -1 + C C = 1 Find the general solution of the equation F’(x) = and
find the particular solution given the point F(1) = 0.
Now plug in (1,0) and solve for C.
0 = -1 + C
Final answer.
C = 1

38. Basic Integration Rules
These two equations allow you to obtain integration formulas directly from differentiation formulas, as shown in the following summary.

39. Basic Integration Rules
cont’d

41. Practice Exercises
Original Rewrite Integrate Simplify

42. Computing Antiderivatives
Problem
Solution

43. Analyzing the motion of an object using antiderivatives
Problem
A particle is moving with the given data. Find the position of the particle.
a(t) = t -3t2, v(0) = 2, s(0) = 5
v(t) is the antiderivative of a(t): v’(t) = a(t) = t -3t2
Antidifferentiation gives v(t) = 10t + 1.5t2 – t3 + C
v(0) = 2 implies that C=2; thus, v(t) = 10t + 1.5t2 – t3 + 2
s(t) is the antiderivative of v(t): s’(t) = v(t) = 10t + 1.5t2 – t3 + 2
Antidifferentiation gives s(t) = 5t t3 – 0.25t4 + 2t + D
s(0) = 5 implies that D=5; thus, s(t) = 5t t3 – 0.25t4 + 2t + 5
Solution

44. Some indefinite integrals