Definite integral of the rate of change of a quantity over an interval interpreted as the change of the quantity over the interval - AP Calculus AB

lim as n approaches infiniti of ((4n3) – 6n)/((n3) – 2n2 + 6)

Use L'Hopitals rule to find the limit.

lim as n approaches infiniti of ((4n3) – 6n)/((n3) – 2n2 + 6)

lim as n approaches infiniti of ((12n2) – 6)/((3n2) – 4n + 6)

lim as n approaches infiniti of 24n/(6n – 4)

lim as n approaches infiniti of 24/6

The limit approaches 4.

The answer is 8.

$y-6x-x^{2}$=4

$y=x^{2}$+6x+4

y'=2x+6

We see the answer is 0 after we do the quotient rule.

h(x)=\frac{g(x)}{(1+f(x))}$

$h'(x)=\frac{g'(x)(1+f(x))-g(x)(f'(x))}{(1+f(x))^{2}$$}

The answer is seconds.

$p=3t^{2}$-t+16

v=p'=6t-1

now set because that is what the question is asking for.

v=0=6t-1

t=\frac{1}{6}$ seconds

The answer is at 6 seconds.

$p=-t^{2}$+12t+2

We can see that this equation will look like a upside down parabola so we know there will be only one maximum.

v=p'=-2t+12

Now we set to find the local maximum.

v=0=-2t+12

t=6 seconds

According to the Fundamental Theorem of Calculus, if we take the derivative of the integral of a function, the result is the original function. This is because differentiation and integration are inverse operations.

For example, if h(x)=int_${a}^{x}$f(u)du, where is a constant, then h'(x)=f(x).

We will apply the same principle to this problem. Because the integral is evaluated from 0 to $x^{2}$, we must apply the chain rule.

g'(x)=\frac{d}{dx}$int_${0}^{x^{2}$$}f(t)dt=f(x^{2}$)cdot $$\frac{d}{dx}$(x^{2}$)

$=2xf(x^{2}$)

The answer is $2xf(x^{2}$).

${$\sqrt{x^2$-9}$}>0 because the denominator cannot be zero and square roots cannot be taken of negative numbers

$x^2$-9>0

$x^2$>9

$$\sqrt{x^2$}$>$\sqrt{9}$

left | x right |>3

x>3: or, x<-3

First, mulitply both sides by r.

Then, use the identities and .

The answer is .

The average function value is given by the following formula:

, evaluated from to .