4.4-4.6 AP Calculus Quiz

Absolute value functions in integration

-find where the function is equal to 0 because this is where the function changes -then split it into two with the area to the left being the function negated and the one to right is the regular function -finally add the answers together because of the additive interval property

(4.6 Begins) Trapezoidal Rule

∫f(x)dx= (b-a)/ 2n [f(x₀) + 2f(x₂) + 2f(x₃)..... f(x)]

Even and Odd Functions w/ Substitution

A function f is even if f (x)= f (x). It is odd if f (-x)=-f(x) 1. If f is even, then ∫f(x) on interval [-a,a] =2∫f(x) on [0,a] 2. If f is odd, the ∫f(x) on interval [-a,a]=0

The Second Fundamental Theorem of Calculus

Assume that f(x) is continuous on an open interval I containing a. Then the area function: A(x)=∫f(t)dt is an antiderivative of f(x) on I; that is, A'(x) = f(x). Equivalently, (d/dx)∫f(t)dt=f(x) *x is the upper limit for all of this* Conditions for this rule: f(t) is a derivative of an integral Derivative matches upper limit of integration Lower limit of integration is a constant.

(4.4 Begins) The First Fundamental Theorem of Calculus

If f is continuous on the interval [a,b] and F is any function that satisfies F '(x) = f(x) throughout this interval then ∫f(x)dx= F(b)-F(a)

Mean Value Theorem

Let F be a function that satisfies the following hypotheses: F is continuous on the closed interval [a,b] F is differentiable on the open interval (a,b) Then there is a number c in (a,b) such that: ∫f(x)dx=f(c)(b-a) or F(b)-F(a)=f(c)(b-a)

The Definite Integral as a Function of x

Let f be a continuous function on [a, b] and x varies between a and b. If x varies, the following is a function of x denoted by g(x): g(x)=∫f(t)dt It should be clear there is an inverse relationship between the derivative and the integral. Thus, the derivative of the integral function is simply the original function.

(4.5 Begins) Integration by Substitution

Steps: 1. Choose inside or one of trig functions to be "u" 2. Derive, du=____dx 3. Integrate u so ∫udx= (u^2)/2 *this depends on properties of original f(x) for sinx it would be sinu * 4. Plug in original "u" to integrated function 5. Multiply your finished answer by (1/coefficient of _____) found from step 2. 6. Don't forget to add +C

Net Change as the Integral of a Rate

The net change in s(t) over an interval [t1,t2] is given by the integral:

The Upper Limit of the Integral is a Function of x

Use the First Fundamental Theorem of Calculus to evaluate the integral: (g(x) is the upper limit and a is the lower limit) ∫f(t)dt = F(g(x))-F(a) Find the derivative of the result: F'(g(x))-F'(a) a is a constant so F'(g(x))×g'(x)-0 f(g(x))×g'(x)

Composite Functions and The Second Fundamental Theorem of Calculus (aka when the conditions aren't met)

When the upper limit of the integral is a function of x rather than x itself, We can use the Second Fundamental Theorem of Calculus together with the Chain Rule to differentiate the integral. think of it like when we use just x you're multiplying by its derivative but thats one so we dont really see it but with this you do see the derivative so you just tack it on. i am an effing god peace out

Determining Average Value

determine f(c) using mvt and then use this to find c