Calc BC Jeopardy, Jan. 3

Review

100

Name the theorem: A function f(x) that is continuous on [a, b] takes on every y-value between f(a) and f(b).

Name the theorem: A function f(x) that is continuous on [a, b] takes on every y-value between f(a) and f(b).

Intermediate Value Theorem

200

d/dx[arccos(x)] =

d/dx[arccos(x)] = -1 / sqrt(1 - x²)

300

lim_{x-> 2}([x² + x - 6] / [x² - 2x])

lim_{x-> 2}([x² + x - 6] / [x² - 2x]) = 5/2 (L'Hopital's Rule)

400

lim_x->infinity(x² - (ln(x))²) / (x + ln(x))

lim_x->infinity(x² - (ln(x))²) / (x + ln(x)) = infinity (L'Hopital 2x)

500

d/dx[cot^-1(x)] =

d/dx[cot-1(x)] = -1 / (x² + 1)

Integration Formulas

100

indefinite integral of sec(x)tan(x)dx =

indefinite integral of sec(x)tan(x)dx = sec(x) + c

200

indefinite integral of csc(x)cot(x)dx =

indefinite integral of csc(x)cot(x)dx = -csc(x) + c

300

indefinite integral of 1/x dx =

indefinite integral of 1/x dx = ln|x| + c

400

indefinite integral of dx / sqrt(a² - x²) =

indefinite integral of dx / sqrt(a² - x²) = sin^-1(x/a) + c 

or -cos^-1(x/a) + c

500

indefinite integral of dx / x*sqrt(x² - 1) =

indefinite integral of dx / x*sqrt(x² - 1) = sec^-1(x) + c

or -csc^-1(x) + c

Integration

100

indefinite integral of xe^{x^2}dx

indefinite integral of xe^{x^2}dx

u = x²   du = 2xdx   xdx = 1/2du 

integral(e^udu) = 1/2e^u + c = 1/2e^{x^2} + c

200

Evaluate the definite integral (1 + 2x)e^-xdx from 0 to infinity

Evaluate the definite integral (1 + 2x)e^-xdx from 0 to infinity

3 (converges)

300

Evaluate the definite integral 1 / (10 + 2z)dz from -5 to 1.

Evaluate the definite integral 1 / (10 + 2z)dz from -5 to 1.

=1/2ln|12| + infinity = infinity --> diverges

400

indefinite integral of w²sin(10w)dw

indefinite integral of w²sin(10w)dw

-w²/10*cos(10w) + w/50*sin(10w) + 1/500*cos(10w) + c

500

indefinite integral of x⁴e^x/2dx

indefinite integral of x⁴e^x/2dx

2x⁴e^(x/2) - 16x³e^(x/2) + 96x²e^(x/2) - 384xe^(x/2) + 768e^(x/2) + c