2 x 2 x 3 x 3
2b
20/30 = (2 x 10)/(3 x 10) = 2/3
(X Squared times a) times b
By using the fundamental theorem of Calculus. i.e. The integral of f(x) = F(x), your limits are [a,b]. Solve: F(b) - F(a). The FTC, second part, says that if f is a continuous real valued function of [a,b] then the integral from a to b of f(x)= F(b) - F(a) where F is any antiderivative of f, that is, a function such that F'(x) = f(x). Example: Evaluate the integral form -2 to 3 of x^2. The integral form -2 to 3 of x^2 = F(-2) - F(3) = -2^3/3 - 3^3/3 = -8/3 - 27/3 = -35/3
The expression 3b x b can be simplified by multiplying the coefficients (numbers) and the variables separately. In this case, 3b x b simplifies to 3b^2, where b^2 represents b squared. This means you multiply the coefficient 3 by the coefficient 1 (since b can be thought of as b^1) to get 3, and you multiply the variable b by itself to get b^2.
2 x 2 x 3 x 3
x^2-x-6
2 x 2 x 2 x 3 x 3 = 72 2^3 x 3^2 = 72
y = 1/2 x - 3/2
Factor them. 2 x 2 x b x b = 4b2 2 x 3 x b x b x b = 6b3 Combine the factors, eliminating duplicates. 2 x 2 x 3 x b x b x b = 12b3, the LCM
2b
(ax + b)^3 = a^3*x^3 + 3*a^2*x^2*b + 3*a*x*b^2 + b^3. Sorry, but it is so clumsy doing this without superscripts!
Explanation: The difference of squares identity can be written: a 2 − b 2 = ( a − b ) ( a b ) The difference of cubes identity can be written: a 3 − b 3 = ( a − b ) ( a 2 a b b 2 ) The sum of cubes identity can be written: a 3 b 3 = ( a b ) ( a 2 − a b b 2 ) So: x 6 − y 6 = ( x 3 ) 2 − ( y 3 ) 2 = ( x 3 − y 3 ) ( x 3 y 3 ) = ( x − y ) ( x 2 x y y 2 ) ( x y ) ( x 2 − x y y 2 ) If we allow Complex coefficients, then this reduces into linear factors: = ( x − y ) ( x − ω y ) ( x − ω 2 y ) ( x y ) ( x ω y ) ( x ω 2 y ) where ω = − 1 2 √ 3 2 i = cos ( 2 π 3 ) sin ( 2 π 3 ) i is the primitive Complex cube root of 1 .
b represents a number ^ represents raised to a power (x - b)(x^2+ bx +b^2) For example: (X^3 - 27) (x - 3)(x^2 + 3x + 3^2) = (x - 3)(x^2 + 3x + 9)
5
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