Can you tell me why

1. To find the area under a curve bounded by 2 x-values, I integrated the curve and substituted the x-values. The area came out -ve.
2. To find the area under a curve bounded by 2 x-values, I integrated the curve and substituted the x-values. The area came out zero, 0.
3. Why is the integral of e^-x = -e^-x and not e^-x?
4. Why is the integral of e^9x = 9e^9x and not e^9x?
5. How can this be solved for x 3 x^2 - 12 = 0?
6. How can this be solved for x 3/2 x^2 - 12/x = 0?

Area under curve

This section is for area under curve questions

Integration of sin x and cos x

This section is for integration questions on sin x and cos x

Integration of polynomial

This section is for integration of polynomial

Differentiation of ln x, e^x, sin x, cos x

This section is for rate of change of ln x, e^x, sin x, cos x questions

Differentiation of polynomial

This section is for rate of change of real life polynomial questions

First Principle derivative

This section is for first principle (delta proceess) questions

Revision questions for matrices

This section is for matrices exam like questions

What is the answer for?

Problem 1

Only answer True when the statement is ALWAYS True.
The function f(x) = e power of x / (x squared − 1) is continuous on [2, 5].

Problem 2
If log(x) = 4 then log(x^2) is ....

Problem 3
log base 2 (3x) + log base 2 (2x) = 3, what is x?

Test the application of derivatives

A small company makes x hand-painted tiles daily at a cost of C(x) = 125 + 30x + 2x^3/2
dollars. What daily production level minimizes the average cost—i.e., the cost per tile?

K.E. = 1/2 m v squared

If you was the scientist to discover the K.E. formula starting from the fact that K.E. is proportional to the velocity to a power by the mass.
Illustrate your steps to lead to the above formula.

What will this give as a graph?

Voltage (V) Current (mA)
0 0
1 20
2 30
3 65
4 98
5 174
6 280

Your Strategy

All students will act as the teacher. Each of you will comment on the strategy
that you will use to teach someone about differentiation of polynomials, logs and ln.
All the common errors you have observed can now be avoided, how?
What is your strategy in your teaching to avoid these errors?

Are these solutions correct?

Solution 1

y = log base 3 (8x^7 + x)
u = 8 x^7 + x
dy/dx = 56 x^6 + 1
= (56 x^6 + 1) / (8 x^7 + x) log base 3 e

Solution 2
P = ln 6s^3 + s
u = 6s^3
dy/dx = 18 s^2
= (18s^2)/(6s^3) + 1

Solution 2
P = ln 6s^3 + s
u = 6s^3 + s
dy/dx = 18 s^2 + 1
= (18s^2 + 1)/(6s^3 + s)

Solution 3
y = 15 x^3 + 7/(2x^3) - x
y = 15x3 + 7 *2 x^-3 - x
dy/dx = 45 x^2 + 42 x^-4 - 1

Solution 3
y = 15 x^3 + 7/(2x^3) - x
dy/dx = 45 x^2 + 7/6 x^-2 - 1

Log checking

Problem 1
log (2x^2 + 6x) - 6 = log (2x)

What can I do now?


Problem 2
Am I going correct?

log (x + 6) = 2 - log (4x)
log (x + 6) = log 100 - log (4x)

Application of simple differentiation

Question 1
Water is being drained from a pond such that the volume V (in m^3) of water in the pond after t hours is given by V = 5000(60 - t)^2. Find the rate at which the pond is being drained after 4 h.

Question 2
The velocity of an object moving with constant acceleration can be found from the equation v = (v[0] ^2 + 2as)^2, where v[0] is the initial velocity, a is the acceleration, and s is the distance traveled. Find dv/ds.

Example 3
The electric field E at a distance r from a point charge is E = k/r^2, where k is a constant. Find an expression for the instantaneous rate of change of the electric field with respect to r.

Example 4
The distance s (in m) traveled by a subway train after the brakes are applied is given by s = 20t - 2t^2. How far does it travel, after the brakes are applied, in coming to a stop?

Examining the accuracy of logs

Look at the following and made some comments to assist. Remember the log base is in [] brackets

Example 1
log [8] (x + 2) = 2 - log [8] 2
log [8] 1 + log [8] 2 = 2 / (x + 2)

Example 2
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) + log [8] 2 = 2
log (2x + 4) = 2
log (2x + 8) = 2
log 8
2x + 4 = 1.8
2x = 1.8 - 4

Example 3
log [8] (x + 2) = 2 - log [8] 2
log [8] x + log [8] 2 = 2 - log [8] 2
log [8] x + 2 log [8] 2 = 2

Example 4
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) = 1.67
log [8] 1.67 = x + 2

Example 5
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) + log [8] 2 = 2
log [8] 2(x + 2) = 2
log [8] (2x + 4) = 2
log [8] 2x = 2 - 4

Example 6
log [8] (x + 2) = 2 - log [8] 2
log (x + 2) / log 8 = 2 - log 2 / log 8
log (x + 2) / 0.903 = 2 - 0.33
log (x + 2) 0.903 * 1.667
log x + 2 = 1.5
log x = 1.5 - 2
log x = 0.5
x = 0.32

The use of ( )

What is the use of a bracket?
4 (x + 3)
4 (x + 3) (x + 3)
4 (x + 3)squared

When can you open out a bracket and when is it not allowed?

What is the differences in the following?
log 4
log 4x
log 4x + 3
log (4x + 3)
log (4x + 3)squared

Simple Algebraic Expressions

What will you say to students who do the following in an attempt to help them?


Example 1
1.14 x + 3.42 = 2x - 1
2x - 1.14x - 1 - 3.42 = 0

Example 2
1.338x - 0.699 = 0.778x + 2.334
1.338x + 0.778x = 2.334 - 0.699

Exponents

Below show some exponent approaches. This time the power will be in brackets i.e. ( )

Example 1
5 (2x - 1) = 6 (x + 3)
Multiply left by 6
Multiply right by 5
30 (2x - 1) = 30 (x + 3)
2x -1 = x + 3

Example 2
5 (2x - 1) = 6 (x +3)
log (5) 2x - 1 = log (6) x + 3
log 2x - 1 / log 5 = log x + 3 / log 6
12x - 6 = 5x + 15

Example 3
5 (2x - 1) = 6 (x +3)
log (5) 2x - 1 = log (6) x + 3
2x - 1 log 5 = x + 3 log 6
2x - 1 log 5 - x + 3 log 6 = 0
log [5] (2x - 1) + log [6] (-x + 3) = 0
log 10x - 5 + 9 - x = 0
log 9x + 9 = 0
log 9x = 9