MTH 161 Test 4 review

evaluate the integral integral from 0 to pi of 19sin x - 13 cos x dx

(-19 cos pi - 13 sin pi) - (-19 cos 0 - 13 sin 0)

d/dx of the integral from 2 to 1/x^2 of (2+sin t) dt

(2+sin(1/x^2))(-2/x^3) *if there is more than just an x value, you multiply what you get through FTC Part 1 by the derivative of the x part

evaluate the indefinite integral integral of (3x-3)^20 dx

(3x-3)^21 divided by 63 + C

integral of x(4+x^2)^10 dx, u = 4+x^2

(4+x^2)^11 / 22 + C

use part 1 of the fundamental theorem of calculus to find the derivative of the function g(s) = integral from 8 to s of (t-t^7)^3 dt

(s-s^7)^3

evaluate the indefinite integral integral of x^2 (x^3+5)^10 dx

(x^3+5)^11 divided by 33 + C

how to solve problems using the FTC part 1

*if the upper limit of the integral is just an x, put the x into the function *if x is the lower limit, take the negative integral of the function and put x into it *if there is more than just an x in either the upper or lower limits, you take the part from the FTC part 1 (by putting x into the function) and then you multiply it by the derivative of the x something

use part 1 of the fundamental theorem of calculus to find the derivative of the function g(x) = integral from 4x to 7x of (u^2-5)/(u^2+5) du

*split into two integrals -4((4x)^2-5)/(4x)^2+5) + 7((7x)^2 - 5) / (7x)^2 + 5)

evaluate the indefinite integral integral of x/(x^2+6)^2 dx

-1/2(x^2+6) + C

evaluate the indefinite integral integral of sin 35 t sec^2 (cos 35t) dt

-1/35 tan (cos(35t))

evaluate the indefinite integral integral of sin 4 pi t dt

-1/4 pi cos (4pi t) + C

integral of (2-x)^6 dx

-1/7 (2-x)^7 + C

evaluate the integral integral from 0 to 2 of (2x-9)(8x^2+4) dx

-184

integral of square root of x sin(1+x^3/2) dx

-2/3 cos (1+x^3/2) + C

evaluate the integral integral from -2 to 0 of 1/2t^7 + 1/5t^6 - t dt

-362/35

evaluate the integral integral from -2 to 3 of (x-4 absolute value of x) dx

-47/2

integral sin x dx

-cos x + C

use part 1 of the fundamental theorem of calculus to find the derivative of the function y = integral from sin x to cos x of (5+v^5)^9 dv

-cos x(5+(sin x^5)^9 - sin x (5+cos (x)^5)^9

integral of cos^4 x sin x dx, u = cos x

-cos^5 x/5 + C

integral csc^2 dx

-cot x + C

integral csc x cot x dx

-csc x + C

the integral of -1 to 2 of x^-2 dx

-x^-1 = -1/x evaluated from -1 to 2 doesn't work, not continuous

integral from -pi/2 to pi/2 of x^2 sin x divided by 1 + x^6 dx

0

find the area of the region enclosed by y = sin x and y = sin2x, x = 0, x = pi/2

1/2

integral from 0 to square root of pi of x cos(x^2) dx

1/2 sin(pi) - 1/2sin(0) = 0

evaluate the indefinite integral integral of sec(32 x) tan(32x) dx

1/32 sec (32 x) + C

evaluate the indefinite integral integral of cos x sin^4 x dx

1/5 sin ^5 x + C

find the area of the region between y = 4x-x^2 and y = 3x

1/6

find the area of the region enclosed by y = x and y = x^2

1/6

average value of a function

1/b-a times the integral from a to b of f(x) dx

find the general indefinite integral integral of 5(sin 2x / sin x) dx

10 sin x + C

integral from 0 to 4 of x divided by square root of 1 + 2x dx

10/3

evaluate the integral integral from 9 to 16 of (6x-6) divided by square root of x

136

`find the area of the region between y = x+2. y = 10-x^2, x = -1, x = 2

19.5

evaluate the integral integral from 0 to 2pi of (1 - absolute value of sin x) dx

2 pi - 4

evaluate the integral by making the given substitution integral of x^2 square root x^3 + 21 dx u = x^3 + 21

2/9 (x^3+21)^3/2 + C

evaluate the integral integral from 0 to 3pi/2 of 7 absolute value sin x dx

21

use part 1 of the fundamental theorem of calculus to find the derivative of the function h(x) = integral from 3 to x^2 of the square root of 5 + r^3 dr

2x square root of 5 + x^6

d/dx of the integral from 0 to 3pi/2 of the absolute value of sin x dx

3

integral of cube root x dx

3/4 x^(4/3) + C

use part 1 of the fundamental theorem of calculus to find the derivative of the function g(x) = integral from 1 to x of 3 divided by t^3 + 5 dt

3/x^3 + 5

find the area of the region between y = 4x, y = x^2

32/3

find the area of the region between x = y^2-4y and x = 3y-y^2

343/24

find the general indefinite integral integral of 4x^2 + 5x^-2

4/3 x^3 - 5x^-1 + C

find the average value of f(x) = 2 cos x on the interval [-pi, pi/2]

4/3(pi)

evaluate the integral integral of 5x(2+x^2)^8 dx

5(2+x^2)^9 + C

find the area between y = 18-x^2, y = x^2 - 14

512/3

evaluate the integral integral from 0 to 1 of (6+2x square root x) dx

6.8

find the area of the region enclosed by 4x + y^2 = 12 and x = y

64/3

integral from -3 to 3 (x^4-5x^2+1) dx

66/5

evaluate the integral integral from 0 to pi/4 of 7sec^2 t dt

7

find the area between y = x^2, y = 7x-6x^2

7/6

evaluate the integral integral from 0 to pi/4 of (8+5 cos^2 x) divided by cos^2 x

8 + 5pi/4

evaluate the integral integral from 0 to 2p/3 of (3sin beta + 3 sin beta tan^2 beta) divided by sec^2 beta

9/2

let R be the region enclosed by the curves x = 2y^2 and y + x = 1 find the area of R

9/8

evaluate the integral integral from pi/4 to pi/3 of 9 sec beta tan beta dbeta

9sec(pi/3) - 9sec(pi/4)

area between curves definition 3

A = integral from c to d of (Xr -Xl) dy

the integral of 0 to 2 of (1+3y-y^2)dy

F(x) = y + 3/2y^2 - y^3/3 evaluated from 0 to 2 16/3

indefinite integral

an indefinite integral is an antiderivative of a function f the integral of f(x) dx = F(x) + C, where F is such that F'(x) = f(x) and C is a constant

d/dx of the integral from 1 to x of (square root of 1 + t^3) dt

square root of 1 + x^3

how to find the distance traveled

take the absolute value of the velocity function

how to find the displacement function when given velocity and a given time interval

take the antiderivative and evaluate it between the endpoints of the function (the start and end points)

how to solve problems using the fundamental theorem of calculus part II

take the antiderivative of the function evaluate it between the two endpoints

integral sec^2 x dx

tan x + C

integral of tan^2 x dx

tan x - x + C *identities

area between curves definition 1

the area A bounded by the curves x = a, x = b, y = f(x), y = g(x) where f and g are continuous and f(x) is greater than or equal to g(x) on [a,b] is A = integral from a to b of [f(x)-g(x)] dx

area between curves definition 2

the area between the curves y = f(x), y = g(x), and between x = a and x = b is A = integral from a to b of the absolute value of f(x) - g(x) dx

net change theorem

the integral from a to b of F'(x) dx = F(b) - F(a)

if f is odd....

then integral from -a to a of f(x) dx = 0

if f is even....

then integral from -a to a of f(x) dx = 2 times the integral from 0 to a of f(x) dx

how to find the displacement function when given velocity

to find the displacement function, take the antiderivative of the velocity function and add C

integral from 0 to pi of tan^2 x dx

unable to complete not continuous at pi/2

integral x^n dx

x^(n+1)/(n+1) +C where n does not equal -1

evaluate the integral integral of sin x / cos^2 x dx

sec x + C

integral sec x tan x dx

sec x + C

tan^2 x

sec^2 x - 1

use part 1 of the fundamental theorem of calculus to find the derivative of the function y = integral from 3 to tan x of square root of 3t + square root of t dt

sec^2(x) square root of 3(tan x ) + square root of tan x

integral cos x dx

sin x + C

the substitution rule

if u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then the integral of f(g(x))g'(x) dx = integral of f(u)du

integral f(x) + g(x) dx

integral f(x)dx + integral g(x)dx

integral f(x) - g(x) dx

integral f(x)dx - integral g(x)dx

integral of x(1+2x^4) dx

integral of (x+2x^5) dx x^2/2 + 1/3x^6 + C

Let v(t) = t^2 - 2t - 8 1 is less than or equal to t is less than or equal to 6 a. find the displacement function b. the displacement during the time period 1 is less than or equal to t is less than or equal to 6 c. the distance traveled during this time period

a. t^3/3 - t^2 - 8t + C b. -10/3 c. 98/3

let g(x) = the integral from 0 to x of f(t) dt, where f is the function whose graph is shown a. evaluate g(0), g(3), g(6).... b. on what interval is g increasing? c. where does g have a maximum value?

a. find the area under the curve between 0 and g(?) b. any time the graph of f is above the x axis c. where the graph of f crosses the x axis, from positive to negative y values

if g(x) = the integral from a to x f(x) dt,,, a) what is g(x), g(-x)? b) does g'(x) and g''(x) exist? c) max/min values d) points of inflection

a) find the area under the curve from a to x of the function b) g'(x) = f(x), so look for f(x) to find g'(x) g''(x) = f'(x), so look at the slope of f at x c) max where graph crosses from above x axis to below, min where graph crosses from below x axis to above d) where the slope of f changes sing

substitution rule for definite integrals

if g' is continuous on [a,b] and f is continuous on the range of u=g(x), then integral from a to b of f'(g(x))g'(x) dx = integral of g(a) to g(b) of f(u) du

integral of cf(x) dx

c times the integral f(x) dx

d/dx of the integral from x to 10 tan x dx

d/dx [-integral from 10 to x tan x dx] -tan x

how to find the distance traveled when given a time interval

determine where the velocity is equal to 0 make a number line using the endpoints of the time interval and the values at which velocity equals 0 in the interval determine if the velocity is positive or negative between those intervals if positive, take the antiderivative of the velocity function and evaluate it between the times at which the velocity is positive if negative, take the antiderivative of the negative velocity function and evaluate it between the times at which the velocity is negative you most likely will have to add multiple sections up at the end

find the average value of f(x) = x - x^2 on [0,2]

f average = 1/2 times the integral of x-x^2 from 0 to 2 1/2 [x^2/2 - x^3/3] evaluated from 0 to 2 1/2 [2- 8/3] = -1/3

let f(x) = square root of x on [0,4] find the average of f over the given interval

f average = 1/b-a times the integral from a to b of f(x)dx f average = 1/4 times the integral from 0 to 4 of x^1/2 1/4[2/3x^3/2] evaluated between 0 and 4 1/4(16/3) = 16/12 or 4/3

odd symmetry

f(-x) = -f(x) symmetric with respect to the origin

even symmetry

f(-x) = f(x) symmetric with respect to the y axis

let f(x) = square root of x on [0,4] find c such that f average = f(c)

f(c) = 4/3 square root of x = 4/3 x = 16/9 c = 16/9

integral

find the antiderivative

how to find a general indefinite integral

find the antiderivative and add C

d/dx

find the derivative

the integral from -pi to pi of f(x) dx where f(x) = x if -pi < or equal to x is < or equal to 0 f(x) = sin x if 0 < or equal to x < or equal to pi

have to break this into two integrals to evaluate the integral from -pi to 0 of x dx + the integral from 0 to pi of sin x dx x^2/2 evaluated from -pi to 0 + -cos x evaluated from 0 to pi -pi^2/2 + 2 2 - pi^2/2 *must determine continuity before doing work, do this by taking the limit as x approaches 0, this function is continuous on the interval from -pi to pi

the fundamental theorem of calculus part II

if f is continuous on [a, b] the integral of a to b of f(x) = F(b) - F(a) where F is any antiderivative of f

fundamental theorem of calculus Part 1

if f is continuous on [a,b], then the function g defined by g(x) = integral from a to x of f(t) dt is continuous on [a,b] and differentiable on (a, b) and g'(x) = f(x)