Unit 6

integral sin(x) dx

-cos(x) + C

integral csc^2(x) dx

-cot(x) + C

integral csc(x) cot(x) dx

-csc(x) + C

general solution for inscribed or circumscribed rectangles

1. Draw a rough graph of the function over the interval. 2. Use formula ∆x = b−a/n to determine each subinterval length. 3. Compute x-coordinates of rectangles at either left-end or right-end. 4. Compute the areas of each rectangle (inscribed or circumscribed). 5. Find the summation of the approximated areas of the rectangles.

procedure for u-substitution for indefinite integrals

1. Look for a piece of the function whose derivative is also in the function. If you're not sure what to use, try the denominator or something being raised to a power in the function. 2. Set u equal to that piece of the function and take the derivative with respect to the variable (this will become du) 3. Use your u and du expressions to replace parts of the original integral, and your new integral will be much easier to solve. It may look like this, for example: ∫ u² du 4. In some cases, the du will not be the actual derivative that is found in the original function given in the problem. In this case, you may have to balance out the equation (that is, making du actually equal whatever derivative is found in the original function). For example, if du = 2x dx, but the derivative in the function is only x, you would need to divide by 2 to both sides and end up getting 1/2du = x dx.

tricks for integration

1. factor out/distribute 2. separate fractions (for example, there may be an x² in the denominator) 3. rewrite according to rules

writing a riemann sum in summation notation

1. find width of each rectangle on [a, b] → ∆x = b - a/n 2. find the right endpoint of each rectangle → xκ = a + ∆x × κ 3. plug in xκ into f(x). look at image to write the summation notation. 3. RHS: k values go from 1 to n 4: LHS: k values go from 0 to n - 1

area of trapezoid

A=½(h)(b₁+b₂) *h = length on the x-axis b₁ = y-value of first point b₂ = y-value of second point

integral 0 dx

C

what are two cases in which f(x) does NOT equal F'(x)? (the original derivative function does not equal the derivative of the antiderivative!)

CASE 1: lower bound is a constant, upper bound is a function of x CASE 2: both lower and upper bounds are a function of x

fundamental theorem of calculus part 2

F(b) - F(a) is the total/net change, or displacement in the antiderivative of f over the same interval. F(x) is the antiderivative

the substitution method for indefinite integral

Given: ∫ f(g(x)) (g'(x)) dx STEP 1: Let u = g(x) , then du = g′(x) STEP 2: Find the antiderivative of f STEP 3: Substitute g(x) back for u

definite integral definition

Let f be continuous on [a, b] and f(x) ≥ 0 for all x in the interval. Let [a, b] be partitioned into n subintervals of equal length ∆x = b−a/n. Then, the definite integral of f over a, b is given by:

indefinite integral

The family of all antiderivatives of a function f(x) is the indefinite integral of f with respect to x and is denoted by f(x) dx If F is any function such that F'(x) = f(x), then f(x) dx = F(x) + C, where C is an arbitrary constant (this is the general solution, as there are no specified limits).

substitution in definite integrals

When we integrate over an interval [a, b], the values of a and b are x-values. The use of u-substitution requires a change in [a, b] to [g(a), g(b)]. After identifying u and du, find the new upper and lower bounds by plugging each of them into the function that equals u. Then, rewrite the integral so that it takes ∫ u du as the base and evaluate using the specific limits

integral a^x dx

a^x / ln(a) + C

how to find general anti-derivative

add 1 to the exponent and then divide by that new power. add C at the end if there are no specified limits.

F(x)

anti-derivative

riemann sums give us an...

approximation of the area under the curve

area under the curve

area between the graph and the horizontal x-axis

definite integral notation

b = end point a = start point f(x) = integrand, or derivative function

when taking the integral of tan(x) or cot(x)...

change them to be in terms of sin(x) and cos(x), and then set u equal to the denomenator

The area under f, the derivative function, on some interval [a, b] accumulates at a decreasing rate so F, the antiderivative function, must be _______________________________ on that interval.

concave down

If the area under f, the derivative function, accumulates at an increasing rate, we know that the graph of F, the antiderivative function, must be ________________________________ on some interval.

concave up

f(x)

derivative

semi-circle

divide area of circle (πr²) by 2

integral e^x dx

e^x + C

odd function

f(-x) = - f(x) , symmetric with origin area would be *zero*

even function

f(-x) = f(x) , symmetric with y-axis add up both sides to get area

how to evaluate the integral of an absolute value function

find the zero(s) of the function by setting it equal to 0. then, create create two integrals that are added to each other, in which the lower bound of the first integral is the lower bound of the given problem and upper bound is the zero value; the lower bound of the second integral is the zero value again, and the upper bound is the upper bound of the given problem. evaluate.

fundamental theorem of calculus part 1

if f(x) is continuous on [a, b] then, g(x) = x∫a f(t)dt is continuous on [a, b] and it is differentiable on (a, b), also g'(x) = f(x). g is an antiderivative of f

when taking the integral of sec(x) or csc(x)...

in the case of sec(x) multiply by sec(x)tan(x)/sec(x)tan(x) OR in the case of csc(x) multiply by csc(x)+cot(x)/csc(x)+cot(x). then set u equal to the denomenator

If the area under f, the derivative function, is positive on some interval [a, b], we know that the graph of F, the antiderivative function, must be ____________________________________ on the interval [a, b].

increasing

what happens to the integral when b is less than a (end point is less than start point)?

it has to go backwards, meaning that the integral must turn into negative and b and a are flip flopped.

integral k f(x) dx

k ∫ f(x) dx

the definite integral of a constant, k, is just...

k(b - a)

integral k dx

kx + C

integral of du/u

ln|u| + C

integral 1/x dx

ln|x| + C

complete the square

means to rewrite as (x + ___)² + ____, where you must evaluate (b/2)²

regardless of concavity, RHS will always be...

overestimate for increasing functions underestimate for decreasing functions

left riemann sum (decreasing)

overestimates

right riemann sum (increasing)

overestimates

how do you find C, the constant?

plug in f(x) and x, then solve for C

integral sec(x)tan(x) dx

sec(x) + C

integral cos(x) dx

sin(x) + C

what do you do to the area when part of the graph is under the x-axis?

subtract the area underneath the x-axis from the area above the x-axis.

integral sec^2(x) dx

tan(x) + C

if the derivative function f has an extrema at some point...

the antiderivative g(x) = x∫0 f(t)dt has an inflection point

when the derivative function f changes signs (crosses the x-axis)...

the antiderivative g(x) = x∫0 f(t)dt is an extrema at that point

when the derivative function f is decreasing...

the antiderivative g(x) = x∫0 f(t)dt is concave down

when the derivative function f is increasing...

the antiderivative g(x) = x∫0 f(t)dt is concave up

when the derivative function f is negative...

the antiderivative g(x) = x∫0 f(t)dt is decreasing

when the derivative function f is positive...

the antiderivative g(x) = x∫0 f(t)dt is increasing

what should you not forget when using trapezoidal sums?

the area of a trapezoid; all points are used; the points inside (excluding the endpoints) must be multiplied by *2*

RHS uses all points except for...

the first point

LHS uses all points except for...

the last point

the definite integral can be used to determine...

the net area (displacement) bounded by the x-axis, the function f(x) and vertical lines x = a and x = b

the integral of the velocity function v(t) dt represents the...

the total change in the position over some interval

If r(t) represents the rate at which your pizza changes temperature in a 475°F pizza oven 5 minutes after the pizza had been placed into the oven to cook, the integral of r(t) dt on [5, 20] is...

the total change in the temperature of the pizza, in degrees Fahrenheit, from time t = 5 minutes to t = 20 minutes

regardless of concavity, LHS will always be...

underestimate for increasing functions overestimate for decreasing functions

left riemann sum (increasing)

underestimates

right riemann sum (decreasing)

underestimates

As k takes on different values, the functions graph as _________________________ transformations of one another, each containing the particular point ____________. Even though the lower bound changed is F′(x) = f(t) ? _______________________

vertical (k, 0) yes

to find left-hand or right-hand sums of *rectangles*...

width × [all y-points] *one y-point may not be included according to LHS or RHS estimations

integral [f(x) plus/minus g(x)]

∫ f(x) dx ± ∫ g(x) dx