Definite Integrals

Graphically solving and interval when it is moved up

- ex: 7~9 (f(x) + 3)dx - move function up said number of spots and find area of resulting rectangle underneath - add area of rectangle to given area under the curve

Finding area geometrically

- graph the function only over the interval given - interval given is b - a - use area formula for shape formed

Rules for answering integral problems when given integrals with answers

- if the bounds are switched, make the answer negative - multiply the answer by the coefficient if there is one - change the bounds if the function is shifted left or right exactly how it is written (ex: 1~5f(x - 3)dx becomes -2~2f(x)dx - if function is shifted up or down, add area of new rectangle to answer - use subtraction to find area of a specific integral if necessary - answer is negative if a = 0 and b is negative 0~-4

Answering integral questions with an odd function

- if the interval is -a to a, the answer is 0 (negative and positive areas cancel each other out) - if the interval is -a to a but the function has absolute value bars around it, the negative area is now positive and you have to add the two areas together for final answer - if you're given the answer for 0 to a, but the question asks for -a to 0, the given answer becomes negative because -a to 0 represents the same area as 0 to a, just negative

Overestimate/underestimate for LRAM and RRAM

- increasing function: LRAM is underestimate and RRAM is overestimate - decreasing function: LRAM is overestimate and RRAM is underestimate

Example problem: express area in terms of a and b

- mark off a and b on graph for width - graph given function - plug a and b into function to get your y value for each point where a and b meet the graph of the function - look at shape formed and use corresponding area formula - plug in a and b values according to area formula and simplify

Total area vs. net area

- total area of an integral means you need to make the negative area positive and add it to the positive area - net area means you have to subtract the negative area from the positive area

Recognizing Fundamental Theorem of Calculus

Ex: F(x) = 0~x root(sect)dt; find F'(0.5) - boundaries are a constant and x (constant = a) - problem asks you to find derivative at 0.5 - boundaries are in terms of x, but function is in terms of t *all of these things show that you should take the derivative of the integral by plugging x in for t (and multiplying by derivative of x boundary if necessary) and then evaluating at 0.5 (plugging in x gets constitutes as taking the derivative and gets rid of integral) Solution to above problem: F'(x) = root(secx) F'(0.5) = root(sec(0.5)) = 1.067

Classic AP problem

Ex: g(x) = 0~x f(t)dt (given with graph) - find g(value) by solving integral (ex: g(-2) = 0~-2 f(t)dt - find g'(value) by finding f(value) *g'(x) = f(x)* - find g''(value) by finding slope of line at that value on graph - find minimum and maximum by using number line with x-intercepts/above-below (*treat graph like a first derivative graph*) - include endpoints in closed interval - thus, find inflection points and concave up/down by labeling number line with extrema and slope - to write equation of line tangent at a point: g'(point) = f(point) this value is the slope; find b value by solving integral at that value (ex: g'(-2) = f(-2) = -2, g(-2) = 0~-2 f(t)dt = -2, so y = -2x -2 - to find absolute max/min value of g: use number line to find mins/maxes, solve integral of whatever values are mins/maxes; smallest/greatest area under curve (smallest/biggest interval) is the absolute max value - if you can't find exact area of integral, look at which one has the most positive area

Chart problems with inconsistent x scale

Ex: x values go 0 3 4 7 8 10 (the difference between them is not constant, so you can't use regular LRAM, RRAM, or Trapezoidal rule - for LRAM, find difference between x values and multiply it by leftmost y value (won't use last y value) Ex: (0,3) (3, 7) so you would do 3(3) - for RRAM, find difference between x values and multiply it by rightmost y value (won't use first y value) Ex: (0,3) (3,7) so you would do 3(7) - for trapezoidal rule, you have to find area of each individual trapezoid using area formula (cannot use regular trapezoidal rule) Ex: (0,3) (3,7) (4,8) so you would do 3(0.5)(3 + 7) + 1(0.5)(7 + 8)

Find values of k that satisfy the equation

Example problem: -10 = k~5(2x - 1)dx -10 = (x^2 -x) |k 5 -10 = (25 - 5) - (k^2 - k) -10 = 20 - k^2 + k k^2 - k - 30 = 0 (k - 6)(k + 5) k = 6, -5 - take antiderivative and set equal to value given - plug boundaries into antiderivative (use FTC) - solve equation for k

Mean Value Theorem for Integrals

If f is continuous on [a,b], then at some point c in [a,b] (b - a)(f(c)) = a~b f(x)dx - basically saying that at some point c the area of a rectangle equals the exact area under the curve - usually only point c that actually works - basically, width of rectangle (b - a) times height at point c (f(c)) equals the exact area under the curve - f(c) (the height) is the average value

Trapezoidal rule

Integral (a to b) = (h/2)(f(x)+2f(x)+2f(x)+f(x)) - used to estimate area under the curve - h = (b - a)/n (n = the number of trapezoids) - for chart problems, h is just the difference between the x values - if a chart problem gives an inconsistent x scale, you have to find the area of each trapezoid and add it together

Simpson's method

Integral (a to b) = (h/3)(f(x)+4f(x)+2f(x)+....+2(x)+4f(x)+f(x)) - used to find area under the curve - h = (b - a)/n (n is number of subintervals, but for simpson's rule, it HAS to be even!

Fundamental Theorem of Calculus

a~b f(t)dt = F(b) - F(a) exact area under the curve = anti-derivative evaluated at x minus anti-derivative evaluated at a - use parentheses to separate F(b) and F(a) - make sure antiderivative evaluated at 0 is actually 0 (ex: cos(0) = 1

Additivity ingetral rule

a~b f(x)dx + b~c f(x)dx = a~c f(x)dx - be aware that you can use this for subtraction problems too

Underestimate/overestimate Trapezoidal rule

concave up: trapezoidal rule is overestimate concave down: trapezoidal rule is underestimate

Finding derivative of an integral

dF/dx = d/dx(a~x f(t)dt) = f(x) - f(t)dt = integrand - processes of differentiation and integration are inverses of each other - the derivative of a function from a constant a to x is actually the integrand in terms of the variable x - in other words, replace any variables in the integrand function with the x from the boundaries - DON'T TAKE DERIVATIVE OF THE INTEGRAND, JUST INSERT X *if the x boundary is a function, you have to multiply the function by its derivative after you've plugged it into the integrand

Definite Integral of a continuous function on [a,b]

dx = rectangle width f(x) = rectangle height used to find the exact area under the curve (sum of the areas of the different rectangles) to find the exact area under the curve, multiply f(x) by dx a~b f(x)dx rectangle width = b - a

Average value

f(c) = average value or height of the function average value = 1/(b - a) a~b f(x)dx - basically just multiply 1/(b - a) by the answer of the integral - for word problems, the unit should be a rate unit (ex: degrees / day)

Integrals on a calculator

fnint(f(x), x, a, b)

zero integral rule

if the integral interval is a to a (same number) the answer is 0

Order of integration rule

switching a and b makes answer negative

LRAM and RRAM chart problems

when solving a chart problem using LRAM and RRAM (Riemann sums): LRAM never uses the last y value RRAM never uses the first y value If the problem asks for an underestimate/overestimate, you need to determine if the function is increasing or decreasing and then decide to use either LRAM or RRAM according to the rules