Math Help

Archive for February, 2008

Question: Find area of triangle using definite integrals?

Using definite integrals find the area of a triangle bounded by the x axis and the two lines:

f(x)= -x+6
f(x)= 2x+6

show me how please!

Answer: First, draw the graph. We see that the x-axis goes from -3 to 6 and the top vertex of the triangle is at (0,6)

Now, we have to split up the integral into two. One from -3 to 0, and the other from 0 to 6

So we have
integral from -3 to 0 of 2x + 6 = x^2 + 6x
F(0) – F(-3) = 9

integral from 0 to 6 of -x + 6 = (-1/2)x^2 + 6x
F(6) – F(0) = 18

So, we add to get that the area is 27

Hope this helps.

Multiple Integrals 9: Finding Volume Problems (Tetrahedron)

This section of Math Help online helps with derivatives. 

What is the definition of derivatives?

Derivatives or differential coefficient for a function f(x) at the argument x the limit of the difference quotient 

 

as the increment tens to 0. 

For functions of a single variable if the left and right hand limits exist ands are equal it is the gradient of the curse at x and is the limit of the gradient of the chord joining the point (x , f(x)) and (x+ , f(x+)). 

 

is the limit of as Q approaches P. 

The function of x defined as the is limit for each argument x is the first derivative of y = f(x); it is the rate of change of the value of the function with respect to the independent variable, and is written: 

, f ‘ (x), or D _x f(x) 

while the ratio of differences of which this is the limit is written . 

The process of extracting the derivative is called differentiation. For example the first derivative of ax ⁿ is anx ^n-1 . The second derivative is the first derivative of the first derivative and is written: 

Question: I need Math Help with exponential functions!?

2x^-1/3=6

x^3/2+1=9

√y^5 =32

x^1/2=7

y^-2=9

Answer: 2x^-1/3=6
x^-1/3 = 3
1 / (x^1/3) = 3
x^1/3 = 1/3
(1/3 is the cube root)
cube both sides
x = (1/3)³ = 1/27

x^3/2+1=9
x^3/2 = 8
take both sides to power 2/3
x = 8^2/3 = 64^1/3 = 4

√y^5 =32
is that √(y⁵) =32
y⁵ = 32² = 1024
y = fifth root of 1024 = 4

x^1/2=7
x = 7² = 49

y^(-2) = 9
y² = 1/9
square root both sides
y = ±1/3

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Exponential Function(Math-Log/Exponential)