Calculus Antiderivatives Unit (4.9-)
Use finite approximation to estimate the area under the graph of the function f(x) = 3-x² + 2x between x = -1 and x=3 for: lower sum with two rectangles of equal width; a lower sum with four rectangles; an upper sum with two rectangles; an upper sum with four rectangles
Draw; n=2, ∆x=3-(-1)/2=2, [-1,1],[1,3], f(-1)=1, f(1)=4, both heights zero as seen form picture; n=4, ∆x=3-(-1)/4=1, [-1,0],[0,1],[1,2],[2,3], f(0)=3, f(2)=3, 1st height = 0, 2nd height = 3, 3rd height = 3, 4th height = 0, (0 times 1) + (3 times 1) + (3 times 1) + (0 times 1) = 6; [-1,1], [1,3], f(-1)=1, f(1)=4, f(3)=0, 1st height = 4, 2nd height = 4, (4 times 2)2=16; [-1,0], [0,1], [1,2], [2,3], f(-1) = 1, f(0)=3, f(1) = 4, f(2) = 3, 1st height = 3, 2nd height = 4, 3rd height = 3, 4th height = 4, (3 times 1)2 + 2(4 times 1) = 14
Estimate the area under the graph of f(x) = 4x³ between x=0 and x=8 using: lower sum with two rectangles of equal width; a lower sum with four rectangles; an upper sum with two rectangles; an upper sum with four rectangles
Draw; n=2, ∆x=8-0/2=4, 4(f(0)+f(4)) = 4(0+256)=1024; n=4, ∆x=9-0/4=2, 2(f(0)+f(2)+f(4)+f(6)) = 2(0+32+256+864)=2304; n=2, ∆x=4, 4(f(4)+f(8)) = 4(256+2048)=9216; n=4, ∆x=2, 2(f(2)+f(4)+f(6)+f(8))=2(32+256+864+2048)=6400
Use rectangles to estimate the area under the parabola y = x² from 0 to 1
Draw; n=4 (# of subintervals/rectangles); ∆x=1-0/4 = 1/4; label x-axis (x₀,x₁,x₂,x₃,x₄); area using left endpoints: 1/4(0²+1/4²+1/2²+3/4²)=7/32; area using right endpoints: 1/4(1/4²+1/2²+3/4²+1²)
Estimate the area under the graph of f(x) = 17-x² on [0,4] by dividing [0,4] into four equal subintervals and using right endpoints as sample points
Draw; n=4, ∆x=4-0/4=1; 1(f(1) + f(2) + f(3) + f(4)) = 38
Estimate the area, A, under the graph of f(x) = 2/x on [1,5] by dividing [1,5] into four equal subintervals and using right endpoints.
Draw; n=4, ∆x=5-1/4=1; 1(f(2) + f(3) + f(4) + f(5)) = 2.56
Estimate the area, A, under the graph of f(x) = 3/x on [1,5] by dividing [1,5] into four equal subintervals and using right endpoints
Draw; n=4, ∆x=5-1/4=1; 1(f(2) + f(3) + f(4) + f(5)) = 3.85
Estimate the area under the graph of f(x) = 3sinx between x = 0 and x = π/3 using five approximating rectangles of equal widths and right endpoints
Draw; n=5, ∆x = π/3 - 0/5 = π/15; ∆x(f(x₁) + f(x₂) + f(x₃) + f(x₄) + f(x₅)) = 1.767
Evaluate the integral I₁ = S1 0 √1-x² dx using known areas
Draw; quarter circle; r=1, π(1)²/4 = π/4
Evaluate the definite integral I = S0 -2 (2+√4-x²)dx by interpreting it in terms of known areas
Draw; split into S0 -2 (2)dx + S0 -2 (√4-x²)dx = 2(2) + π(2)²/4 = (4+π)u²
Evaluate: S0 -3 (1+√9-x²)dx
Draw; split into straight line and quarter circle; S0 -3 1dx + S0 -3 (√9-x²)dx; (3 + 9/4πr²)u²
Evaluate: S9 0 (1/3x-2)dx
Draw; split; S6 0(1/3x-2)dx + S9 6(1/3x - 2)dx; 1/2(6)(-2) + 1/2(3)(1) = -6 + 1.5 = -4.5u²
Find the total area of the region between the x-axis and the graph of y = -x²-3x, -4≤x≤1
Draw; zeroes at -3,0; [-4,-3]: negative, [-3,0]: positive, [0,1]: negative; A = |S-3 -4 (-x²-3x)dx| + S0 -3 (-x²-3x)dx| + S1 0 (-x²-3x)dx|; [-x³/3 - 3x²/2]-3 -4 + [-x³/3 - 3x²/2]0 -3 + [-x³/3 - 3x²/2]1 0; |-11/6| + |9/2| + |-11/6| = 49/6
Find the total area of the region between the x-axis and the graph of: y = 2x³ + 12x² + 16x, -4≤x≤0
Draw; zeros at -4,-2,0; [-4, -2], [-2,0]; A = |S-2 -4 (1/2x⁴ + 4x³ + 8x²)dx| + |S0 -2 (1/2x⁴ + 4x³ + 8x²)dx|; [1/2x⁴ + 4x³ + 8x²]-2 -4 + [1/2x⁴ + 4x³ + 8x²]0 -2; |8| + |-8| = 16
A function F is called an antiderivative of f on an interval I if
F'(x) = f(x) for all x in I
Most general antiderivative
F(x) + C
Find the antiderivative: f(x) = sinx; f(x) = 1/x; f(x) = xⁿ, n ≠ -1
F(x) = -cosx + C; F(x) = ln|x| + C; F(x) = xⁿ⁺¹/n+1 + C
F(x) = x⁷, find the antiderivative
F(x) = 1/8x⁸/8 + C
Find the antiderivative: f(x) = 2cosx - 7sinx
F(x) = 2sinx + 7cosx + C
Find the antiderivative: f(x) = 9x²-6x+6
F(x) = 3x³-3x²+6x+C
If F(x) = d/dx(Sx⁴ 0 2t³dt), determine the value of F(1)
F(x) = 4x³(2(x⁴)³) = 8x¹⁵; plug in 1 = 8
F(x) = 9, find antiderivative
F(x) = 9x + C
F(x) = e^x, antiderivative
F(x) = e^x + C
F(x) = 1/x, find antiderivative
F(x) = lnx + C
Find the antiderivative, C = 0: f(x) = 1/x; g(x) = 11/x; h(x) = 5-4/x
F(x) = ln|x|; G(x) = 11ln|x|; H(x) = 5x-4ln|x|
F(x) = cosx, antiderivative
F(x) = sinx + C
Power rule formula for antiderivative
F(x) = xⁿ; antiderivative: f(x) = xⁿ⁺¹/n+1 + C
Find the derivative: g(x) = S1+2x 1-2x tsintdt
Fast way: (1+2x)sin(1-2x)(2) - (1-2x)sin(1-2x)(-2) = (2+4x)sin(1+2x) - (2-4x)sin(1-2x); book way: S0 1-2x tsintdt + S1+2x 0 tsintdt; -S1-2x 0 tsintdt + " then plug in everything
Explain the notation: Sb a f(x)dx
From a to b, left to right
Find the antiderivative: g(x) = 1/x³
G(x) = -1/2x⁻²
Find the antiderivative: g(x) = -5x⁻⁶
G(x) = -5x⁻⁵/-5 = x⁻⁵
Find the antiderivative: h(x) = x⁻⁶
H(x) = -1/5x⁻⁵
Find the antiderivative: h(x) = 7/x³
H(x) = -7/2x⁻²
Second part of the fundamental theorem of calculus
If f is continuous on [a,b] then Sb a f(x)dx = F(b) - F(a); no +C's because cancel out
Find the antiderivative: k(x) = x⁻⁶ + 2x + 4
K(x) = -1/5x⁻⁵ + x² + 4x
Find the antiderivative: k(x) = 6 - 7/x³
K(x) = 6x + 7/2x⁻²
Evaluate the integral: S-1 -2 1/x dx
Ln|x|]-1 -2; ln1 - ln2 = -ln2
What happens if portion under x-axis?
Negative
Evaluate the definite integral I = S4 0 (|x²-4| - x²)dx
Number line to see when negative/positive; S5 0 |x^2-4|dx - S5 0 x^2dx; -|S2 0 (x^2-4)dx| + S5 2 (x^2-4)dx; x^3/3 - 4x]2 0 + x^3/3-4x]5 2; 16/3 + (39-12) - 125/3
Evaluate the definite integral I = S6 3 |x-4|dx
Number line to see when negative/positive; break up into |negative antiderivative| + positive antiderivative; 5/2; see answer key
For decreasing, left is___and right is____
Over; under
Find the derivative of the function: g(x) = S3x 2x (u²-1)/(u²+1)du
Plug in times chain rule: (3x)²-1/(3x)²+1(3) - (2x)²-1/(2x)²+1(3) = 27x²-3/9x²+1 - 8x²-2/4x²+1 = g'(x); book way: d/dxS0 2x u²-1/u²+1dx + d/dxS3x 0 u²-1/u²+1dx; -d/dxS2x 0 u²-1/u²+1 " and plug in everything
Estimate the area under the graph of f(x) = 3sinx between x=0 and x = π/3 using five approximating rectangles of equal widths and right endpoints
Right endpoints: A = ∆x(f(x₁) + f(x₂) + ... + f(xn)); ∆x=b-a/n = π/3-0/2 = π/15; A = π/15(f(π/15) + f(2π/15) + f(π/5) + f(4π/15) + f(π/3)) = 1.767
Evaluate the integral: S -3cosx/sin²x dx
S -3cosx/sinx times 1/sinx = S-3cotxcscx; 3cscx+C
Determine the integral I = S(2x⁴ + 3x² + 5/x²)dx
S 2x² + 3 + 5x⁻² dx = 2/3x³ + 3x - 5x⁻¹ + C
Determine the integral I = S x(4+9x⁴)dx
S 4x + 9x⁵dx = 2x² + 3/2x⁶ + C
Find the indefinite integral: S(18x+8)dx
S(18x+8)dx = 9x²+8x+C
Find the indefinite integral: S(3t²+t/4)dt
S(3t²+t/4)dt = t³+1/8t²+C
Find the general indefinite integral: Sect(sect+tant)dt
S(sec²t + secttant)dt = tan(t) + sec(t) + C
Find the indefinite integral: S(¹¹√x + ¹²√x)dx
S(¹¹√x + ¹²√x)dx=11/12x^12/11 + 12/13x^13/12 + C
Evaluate the integral: S -5/cscx dx
S-5sinx dx; 5cosx + C
Evaluate the definite integral I = S0 -4 (2+√16-x²)dx by interpreting it in terms of known areas
S0 -4 (2)dx + S0 -4 (√16-x²)dx; Draw; 8 + 6π
Evaluate the integral: S1 -1 t(1-t)²dt
S1 -1 (t-2t²+t³); 1/2t² - 2/3t³ + 1/4t⁴]1 -1 = (1/2(1)² - 2/3(1)³ + 1/4(1)⁴) - (1/2(-1)² + 2/3(-1)³ + 1/4(-1)⁴) = -4/3
Find the area under the parabola y = x² from 0 to 1
S1 0 x²dx = x³/3]1 0 = 1³/3 - 0³/3 = 1/3
S2 -2 (1+√4-x²)dx =
S2 - 2 1dx + S2 -2(√4-x²)dx; straight line = 4; semicircle = 2π; A = (4+2π)u²
Evaluate the integral: S2 1 (4+u²/u³)du
S2 1 4u⁻³ + S1/udu = -2u⁻² + ln|u|]2 1 = [-2(1/2²) + ln2] - [-2 + ln1] = 3/2 + ln2
Evaluate the integral: S20x⁸ + 5x³ - 12/x⁵ dx
S20x³ + 5x⁻² - 12x⁻⁵ dx; 5x⁴ - 5/x + 3/x⁴ + C
Find the general indefinite integral: Ssin2x/sinx dx
S2sinxcosx/sinx dx = 2cosx; 2sinx + C; sin2x = 2sinxcosx
Find the indefinite integral: S5x(7-x⁻⁵)dx
S5x(7-x⁻⁵)dx = 35/2x² + 5/3x⁻³ + C
Evaluate S6 3 dx/x
S6 3 (1/x)dx = ln|x|]6 3 = ln|6| - ln|3| = ln6/3 = ln|2|
Determine the integral I = S(6-5x)/√x dx
S6x^-1/2 - 5x^1/2 dx; 12x^1/2 - 10/3x^3/2 + C
Evaluate the integral: S6x⁶ - 5x² - 12/x⁴ dx
S6x² - 5x⁻² - 12x⁻⁴ dx; 2x³ + 5/x + 4/x³ + C
Find the indefinite integral: S8costdt
S8costdt = 8sint + C
Find the indefinite integral: S 8sec²xdx
S8sec²xdx = 8tanx + C
Sa -b |x|dx =
Sa 0 (x)dx + S0 -b (-x)dx
Si(x) h(x) f(t)dt =
Sa h(x) f(t)dt + Si(x) a f(t)dt
Constant function property of definite integrals
Sb a cdx = c(b-a) where c is any constant; straight line
Definite integral
Sb a f(x)dx
Sc a f(x)dx + Sb c f(x)dx =
Sb a f(x)dx
Sb a [f(x) + g(x)]dx =
Sb a f(x)dx + Sb a g(x)dx
Sb a [f(x) - g(x)]dx =
Sb a f(x)dx - Sb a g(x)dx
Formula for displacement
Sb a v(t)dt
A particle moves along a line so that its velocity at time t is v(t) = t² -t - 6 (m/s). Find the displacement of the particle during the time period 1≤t≤4; find the distance traveled during this time period
Sb a v(t)dt; S4 1 (t² - t - 6)dt, 1/3t³ - 1/2t² - 6t]4 1 = (1/3(4)³ - 1/2(4)² - 6(4)) - (1/3(1)³ - 1/2(1)² - 6(1)) = -32/3 + 37/6 = -9/2; draw graph to see which interval negative, S4 1 |t² - t - 6|dt = |S3 1 (t² - t - 6)dt| + S4 3 (t² - t - 6)dt, 1/3t³ - 1/2t² - 6t]3 1 + 1/3t³ - 1/2t² - 6t]4 3, |((1/3(3)³ - 1/2(3)² - 6(3)) - (1/3(1)³ - 1/2(1)² - 6(1)))| + ((1/3(4)² - 1/2(4)² - 6(4)) - ((1/3(3)³ - 1/2(3)² - 6(3))) = 44/6 + 17/6 = 61/6
Formula for distance traveled
Sb a |v(t)|dt
Find the antiderivative: r(θ) = secθtanθ - 2e^θ
Secθ - 2e^θ + C
Double angle formulas
Sin(2x) = 2sinxcosx; cos(2x) = cos^x - sin^2x; tan(2x) = 2tanx/1-tan^2x
How to find the acceleration when given a velocity graph?
Slope
Evaluate the integral I = S3 1 (5+4x)dx by interpreting it in terms of known areas
Split into S3 1 (5)dx + S3 1 (4x)dx = 2(5) + 1/2(2)(4+12) = 26
Find the indefinite integral: Sx^-1/6dx
Sx^-1/6dx = 6/5x^5/6 + C
Notation for antiderivative starting with x²
Sx²dx = ___ + C
S3π/2 0 |sinx|dx
Sπ 0 sin dx + S3π/2 π (-sinx)dx; -cosx]π 0 + cosx]3π/2 π; (-cos(π) - (-cos(0)) + (cos(3π/2) - cos(π)) = 1 + 1 + 1 = 3
Evaluate the integral: Sπ/3 0 ((sinθ+sinθtan²θ)/sec²θ)dθ
Sπ/3 0 sinθ(1+tan²θ)/sec²θ dθ; Sπ/3 0 sinθ(sec²θ)/sec²θ dθ = -cosθ]π/3 0 = -cos(π/3) - (-cos(0)) = -1/2 + 1 = 1/2
Evaluate the integral I = Sπ/6 0 2sin2x/cosx
Sπ/6 0 2sinxcosx/cosx, 2sinx; -2cosx, plug in
Evaluate the integral: S18 1 (√3/z)dz
S√3z^-1/2dt = 2√3z^1/2]18 1 = 2√3(18^1/2 - 1^1/2) = 6√6-2√3
Fundamental theorem of calculus essentially says
The derivative of the antiderivative is the original function; g'(x) = f(x)
For increasing, left is__ and right is____
Under; over
The acceleration function in (m/s²) and the initial velocity are given for a particle moving along a line. Find a) the velocity at time t, and b) the distance traveled during the given time interval: a(t) = t + 4, v(0) = 5, 0≤t≤10
V(t) = 1/2t² + 4t + C, C = 5, v(t) = 1/2t² + 4t + 5; S10 0 (1/2t² + 4t = 5)dt, 1/6t³ + 2t² + 5t]10 0, ((1/6(10)³ + 2(10)² + 5(10)) - (1/6(0)³ + 2(0)² + 5(0)) = 2500/6 m
The velocity function (in m/s) is given for a particle moving along a line. Find a) the displacement b) the distance traveled by the particle during the given time interval: v(t) = t² - 2t - 8, 1≤t≤6
V(t) = 1/3t³ - t² - 8t, F(b) = F(a), (1/3(6)³ - (6)² - 8(6)) - (1/3(1)³ - (1)² - 8(1)) = -10/3m; Sb a |v(t)|dt, S6 1 |t²-2t-8|dt, t = 4, -S4 1 (t² - 2t - 8)dt + S6 4 (t² - 2t - 8)dt = 1/3t³ - t² - 8t]4 1 + 1/3t³ - t² - 8t]6 4 = 98/3
The velocity function (in m/s) is given for a particle moving along a line. Find a) the displacement b) the distance traveled by the particle during the given time interval: v(t) = 3t-5, 0≤t≤3
V(t) = 3/2t² - 5t, F(b) - F(a), (3/2(3)² - 5(3)) - (3/2(0)² - 5(0)) = -3/2 m; Sb a |v(t)|dt, S3 0 |3t-5|dt, 3t-5 = 0, t=5/3, -(3t-5), 3t-5, S5/3 0 -(3t-5)dt + S3 5/3 (3t-5)dt, -3/2t² + 5t]5/3 0 + 3/2t² - 5t]3 5/3, plug in and get 41/6
The acceleration function in (m/s²) and the initial velocity are given for a particle moving along a line. Find a) the velocity at time t, and b) the distance traveled during the given time interval: a(t) = 2t+3, v(0) = -4, 0≤t≤3
V(t) = t² + 3t + C, C=-4, v(t) = t² + 3t - 4; S3 0 |t²+3t-4|dt, -S1 0 t² + 3t -4dt + S3 1 t² + 3t - 4dt, -(1/2t³ + 3/2t² -4t]1 0 + 1/3t³ + 3/2t² - 4t]3 1, -(1/3(1)³ + 3/2(1)² - 4(1) - 0) + ((1/3(3)³ + 3/2(3)² - 4(3)) + (1/3(1)³ + 3/2(1)² - 4(1)) = 89/6m
For graphing, describe curves
Where original graph below x-axis, decreasing; where above x-axis, increasing
S2 -1 (x-2|x|)dx
X-2x = -x; x+2x = 3x; S2 0 (-x) + S0 -1 (3x); -1/2x²]2 0 + 3/2x²]0 -1; (-1/2(2)² - -1/2(0)²) + (3/2(0)² - 3/2(-1)²) = -7/2
Find F'(x): F(x) = Sx 3 t^1/3 dt
X^1/3 - 3√3
Find the antiderivative: f(x) = x^3.4 - 2x^(√2)-1
X^4.4/4.4 - √2x^√2 + C
Find the antiderivative: f(x) = e²
Xe² + C
Find the antiderivative: f(x) = x(2-x)²
X³-4x⁷+4x -> 1/4x⁴-4/3x³ + 2x² + C
Find F'(x): F(x) = Sx² x (-t³ + 3t + 3)dt
[2x(-(x²)³ + 2(x²) + 3)] - [-(x)³ + 3(x) + 3]; -2x⁷ + 7x³ + 3x - 3
Evaluate the integral I = S1 0 (2x - x^1/3)dx
[x - 3/4x^4/3]1 0 = (1-3/4) - 0 = 1/4
A car is at 90mph when its brakes are fully applied, producing a constant deceleration of 22ft/s². Find the distance covered by the car before it comes to a stop. (60mph = 88ft/s)
a(t) = -22; v(t) = -22t + C, v(0)=90mph -> 132ft/s = C, v(t) = -22t + 132, t = 6s; s(t) = -11t² + 132t + D, D=0; plug in 6 = 396 ft
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/s². What is the distance traveled before the car comes to a stop?
a(t) = -22ft/s²; v(t) = -22t + C; v(0) = 50mi/hr times 5280ft/s times hr/3600s = 220/3 ft/s = C; s(t) = -11t² + 220/3t; v(t) = -22t + 220/3t + D; D = 0; 22t = 220/3, t = 10/3s; plug in and get 122.23 ft
A stone is dropped off a cliff and falls under gravity with a constant acceleration of -32 ft/sec². If it hits the ground with a speed of 96 ft/sec, determine the height of the cliff
a(t) = -32 ft/s²; v(t) = -32t + C, v(0)=0, C=0; v(t) = -32t = -96, t=3s; s(t) = -16t²+D, s(3)=0, D=144ft
∆x=
b-a/n
Function, find antiderivative; cf(x); f(x) + g(x); xⁿ (n≠-1); 1/x; e^x; cosx; sinx; sec²x; secxtanx
cF(x); F(x) + G(x); xⁿ⁺¹/n+1; ln|x|; e^x; sinx; -cosx; tanx; secx
Sb a xf(x)dx =
cSb a f(x)dx
Find the curve in the xy-plane that passes through the point (4,8) and whose slope at each point is 6√x
dy/dx = 6√x; 6x^1/2; 4x^3/2 + C, plug in 4 and get y = 4x^3/2 - 24
If the graph of f passes through the point (1,-1) and the slope of the tangent line at (x,f(x)) is 8x-7, find the value of f(2)
dy/dx = 8x-7; f(x) = 4x²-7x+C, C=2; f(x) = 4x²-7x+2, f(2) = 4
Find the derivative: g(x) = Sx 3 e^(t²-t)dt
e^(t²-t)
Find the derivative: F(x) = Sx² x e^t²dt
e^(x²)²(2x) - e^x²(1) = 2xe^x⁴ - e^x²; book way: S0 x e^t²dt + Sx² 0 e^t²dt, Sx²0 e^t²dt - Sx 0 e^t²dt
Se^axdx =
e^ax/a + C
Evaluate: S3 1 e^xdx
e^x]3 1 = e³ - e
Evaluate the integral: S5 -5 edx
ex]5 -5 = 5e - (-5e) = 10e; treat as constant, not e¹
Determine f(t) when f"(t) = 6(2t+1) and f'(1) = 3, f(1) = 4
f"(t) = 12t+6; f'(t) = 6t²+6t+C, f'(1) = 6(1)²+6(1) + C = 3, 12+C = 3, C=9; f'(t) = 6t²+6t+9; f(t) = 2t³+3t²-9t+C, C=8; f(t) = 2t³+3t²-9t + 8
Find the value of f(e) when f"(x) = 4/x², x>0 and f(1) = 3, f'(2)=4
f"(x) = 4x⁻²; f'(x) = -4x⁻¹ + C, C=6; f'(x) = -4x⁻¹ +6; f(x) = -4+6x+C=3, C=1; f(x) = 6e-7
Find f: f"(t) = 2e^t + 2sint, f(0) = 0, f(π) = 0
f'(t) = 2e^t - 3cost + C; f(t) = 2e^t - 3sint + Ct + D, D = -2; plug in π, C = -2e^π/π; f(t) = 2e^t - 3sint - 2e^π/π - 2
Find f if f"(x) = 12x² + 6x - 4, f(0) = 4, and f(1) = 1
f'(x) = 12x³/3 + 6x²/2 - 4x + C; f'(x) = 4x³ + 3x² - 4x + C; f(x) = 4x⁴/⁴ + 3x³/3 - 4x²/2 + Cx + D; f(x) = x⁴ + x³ - 2x² + Cx + D; plug in 0, D = 4, plug in 1 and get C = -3; f(x) = x⁴+x³-2x²-3x+4
Find f: f"(x) = 8x³ + 5, f(1) = 0, f'(1) = 8
f'(x) = 2x⁴ + 5x + C, C = 1; f'(x) = 2x⁴ + 5x + 1; f(x) = 2/5x⁵ + 5/2x² + x + C, C = -39/10; f(x) = 2/5x⁵ + 5/2x² + x - 39/10
Find f(x) on (-π/2,π/2) when f'(x) = 3+tan²x and f(0)=2
f'(x) = 3+(-1+sec²x) = 2+sec²x; f(x) = 2x+tanx + C, C=2; f(x) = 2+2x-tanx
Sb a f(x)dx
f(b) - f(a)
d/dxSh(x) a f(t)at =
f(h(x))h'(x)
Find f: f'(t) = t + 1/t³, t>0, f(1) = 6
f(t) = 1/2t² - 1/2t⁻² + C; C = 6; f(t) = 1/2t² - 1/2t⁻² + 6
d/dxSx a f(t)dt =
f(x)
Find f(x) when f'(x) = 2cosx - 9sinx and f(0)=6
f(x) = 2sinx + 9cosx + C; C=-3; f(x) = 2sinx + 9cosx - 3
Find f such that f'(x) = 8x² + 3x - 5 and f(0) = 1
f(x) = 8/3x³ + 3/2x² - 5x + 1; C = 1
Find f: f'(x) = 5x⁴ - 3x² + 4, f(-1) = 2
f(x) = x⁵-x³+4x+C; C=6; f(x) = x⁵ - x³ + 4x + 6
Find all functions g such that g'(x) = 4sinx + (2x⁵-√x/x)
g'(x) = 4sinx + 2x⁴ - x^-1/2; G(x) = -4cosx + 2x⁵/5 - 2x^1/2 + C
Determine g'(x) when g(x) = Sx 0 √6-5t²dt
g'(x) = √6-5x²; 1st fundamental theorem
Find d/dx Sx⁴ 1 sectdt
g(x) = Sh(x) a f(t)dt; g'(x) = f(h(x))h'(x); g'(x) = sec(x⁴)4x³; g'(x) = 4x³sec(x⁴)
Fundamental theorem of calculus notation
g(x) = Sx a f(t)dt
Find the derivative of the function: g(x) = Sx 0 √1+t² dt
g(x) = Sx a f(t)dt; g'(x) = f(x); g'(x) = √1+x²
Use finite approximation to estimate the area under the graph of f(x) = x² and above the graph of f(x) = 0 from x₀=0 to xₙ=14 using: lower sum with two rectangles of equal width; a lower sum with four rectangles; an upper sum with two rectangles; an upper sum with four rectangles
n=2, ∆x=14-0/2=7, (x₁-x₀)(f(x₀) + (x₂-x₁)(f(x₁)) = 343; n=4, ∆x=14-0/4 = 3.5, (x₁-x₀)f(x₀) + (x₂-x₁)f(x₀) + (x₃-x₂)f(x₂) + (x₄-x₃)f(x₃) = 600.25; (x₁-x₀)f(x₁) + (x₂-x₁)f(x₂) = 1715; (x₁-x₀)f(x₁) + (x₂-x₁)f(x₂) + (x₃-x₂)f(x₃) + (x₄-x₃)f(x₃)=1286.25
Estimate the area under the graph of f(x) = √x from x = 0 to x = 4 using four approximating rectangles and right endpoints. Sketch the graph and the rectangles. Is your estimate an underestimate or overestimate?
n=4, ∆x=1; A=3+√2+√3
Find the position of the particle: v(t) = 1.5√t, s(4) = 10
s(t) = t^3/2 + C; C=2; s(t) = t^3/2 + 2
Evaluate the integral: Sπ/4 0 secθtanθdθ
secθ]π/4 0 = 2√2/2 - 1
Find the position of the particle: a(t) = t²-4t+6, s(0) = 0, s(1) = 20
v(t) = 1/3t³ -2t² + 6t + C, C =0; v(t) = 1/3t³ -2t² + 6t; s(t) = 1/12t⁴ - 2/3t³ + 3t² + C, C = 211/12; s(t) = 1/12t⁴ - 2/3t³ + 3t² + 211/12
Find the position of the particle: a(t) = 3cost - 2sint, s(0) = 0, v(0) = 4
v(t) = 3sint + 2cost + C; C=2; v(t) = 3sint + 2cost + 2; s(t) = -3cost + 2sint + 2t + C, C = 3; s(t) = -3cost + 2sint + 2t + 3
Evaluate the integral: S3 0 (1 + 6w² - 10w⁴)dw
w+2w³-2w⁵]3 0 = (3+2(3)³-2(3)⁵) - (0+2(0)³-2(0)⁵) = -429
Evaluate the integral: S1 0 (1+1/2u⁴ - 2/5u⁹)du
x+5/2u⁵ - 2/50u¹⁰]1 0 = (1+5/2-2/50) - (0) = -73/50
Evaluate: S3 1 (3x²-x⁴/6)dx
x³-1/30x⁵]3 1 = 567/30-29/30 = 269/15
Evaluate: S4 1 (x²+√x)dx
x³/3 + 2x^3/2/3]4 1 = 80/3-1 = 77/3
Evaluate: S3 0 x(x-4)dx
x³/3 - 2x³]3 0 = -9-0 = -9
Evaluate the integral: S1 -1 x¹⁰⁰dx
x¹⁰¹/101]1 -1 = 1/101 - (-1/101) = 2/101
Evaluate the integral: S2 1 (4x³-3x² + 2x)dx
x⁴-x³+x²]2 1 = ((2)⁴-(2)³+2²) - (1⁴-1³+1) = 11
A car heads slowly north from Austin on IH 35. Its velocity t hours after leaving Austin is given (mph) by v(t) = 20 + 19t - 6t². How many miles will the car have covered during the first 2 hours of driving?
|20 + 19t - 6t²|; S2 0 |20 + 19t - 6t²|, always positive; [20 + 19/2t² - 2t³]2 0 = 62 miles
Estimate the area, A, under the graph of f(x) = 5/x, on [1,5] by dividing [1,5] into four equal subintervals and using right endpoints.
∆x=5-1/4=1; A = 1(f(2) + f(3) + f(4) + f(5))=77/12
Find dy/dr for y = √11+12t² dt
√11+12r²
Find the derivative: g(r) = Sr 0 (√x²+4)dx
√x²+4
Find the general indefinite integral: Sv(v²+2)dv
1/6v⁶ + v⁴ + 2v² + C
Evaluate the integral: S2 0 (y-1)(2y+1)dy
2/3y³ - 1/2y² - y]2 0 = 4/3
Find the general indefinite integral: S(√x³+³√x²)dx
2/5x^5/2 + 3/5x^5/3 + C
Evaluate the integral: S4 0 (3√t - 2e^t)dt
2t^3/2 - 2e^t]4 0 = (2(4)^3/2 - 2e⁴) - (2(0)^3/2 - 2e⁰) = 18-2e⁴
Determine the integral I = S(2+x^-5/4)dx
2x - 4x^-1/4 + C
S2 0 |2x-1|dx
2x-1 = 0, x = 1/2; S1/2 0 |2x-1|dx + S2 1/2 |2x-1|dx; S1/2 0 (1-2x)dx + S2 1/2 (2x-1)dx; x-x²]1/2 0 + x² -x]2 1/2; [(1/2-(1/2)²) - (0-0²)] + ((2² -2) - (1/2² - 1/2)) = 5/2
Find the antiderivative: f(t) = 3t⁴ - t³ + 6t²/t⁴
3 - 1/t + 6/t² = 3-1/t + 6t⁻²; 3t - (ln|t|) - 6/t + C
Find F'(x): F(x) = S3x 0 (t³ - 4t² + 6)dt
3((3x)³ - 4(3x)² + 6) - (0³ -4(0)² + 6) = 91x³ - 108x² + 12
Find the antiderivative: f(x) = ³√x² + x√x
3/5x^5/3 + 2/5x^5/2 + C
Evaluate the integral: S8 1 x^-2/3dx
3x^1/3]8 1 = 3(8)^1/3 - 3(1)^1/3 = 3
F(x) = 3x⁹, find antiderivative
3x^9+1/9+1 + C = 3x¹⁰/10 + C
f(x) = 3x², find antiderivative
3x³/3 + C = x³ + C
Evaluate the integral I = S4 0 4√16-x²dx by relating it to the graph of y = √16-x²
4S4 0 √16-x²dx = π(4)²/4 times 4 = 16π
Find the derivative: y = Sx⁴ 0 cos²θdθ
4x³(cos²θ)
Evaluate the integral: S1 0 (5x-5^x)dx
5/2x² - 5^x/ln5]1 0 = (5/2-5/ln5) - (-1/ln5) = 5/2 - 4/ln5
Evaluate the integral: S-1 -3 5/x dx
5ln|x|]-1 -3; -5ln1 - 5ln3 = -5ln3
Find all functions g such that g'(x) = 5x²+4x+5/√x
5x²/x^1/2 + 4x/x^1/2 + 5/x^1/2; 5x^3/2 + 4x^1/2 + 5x^-1/2; g(x) = 2x^5/2 + 8/3x^3/2 + 10x^1/2 + C; g(x) = 2√x(x²+4/3x+5)+C
Evaluate: S2 0 (7x²-4x+4)dx
7x³/3 - 2x²+4x]2 0 = 56/3
Evaluate the integral: S4 0 (4-t)√t dt
8/3t^3/2 - 2/5t^5/2]4 0 = 128/15
Evaluate: Sπ/3 0 8sec²xdx
8tanx]π/3 0 = 8√3
Find the antiderivative: f(x) = 8x⁹ - 3x⁶ + 12x³
8x⁹+1/9+1 - 3x⁶+1/6+1 + 12x³+1/3+1 = 4/5x¹⁰ - 3/7x⁷ + 3x⁴ + C
Find the indefinite integral: S(-7secxtanx - 5sec²x)dx
= -7secx - 5tanx + C
Find the indefinite integral: S -cscθcotθ/5 dθ
= 1/5cscθ + C
Find the indefinite integral: S 4x^√7 dx
= 4x^√7 + 1/√7 + 1 + C
Find the indefinite integral: S(t⁵√t² + √t/t²)dt
= 5/2t^2/5 -2t^-1/2 + C
Area of a trapezoid
A=1/2h(b₁+b₂)
Area using left endpoints
A=∆x(f(x₀) + f(x₁) +...+f(xₙ-₁)
Area using right endpoints
A=∆x(f(x₁) + f(x₂)+...+f(xₙ)
When given graph, how to tell distance (quest #9-12)
Area of interval
How do you tell which graph is antiderivate
Below x-axis, decreasing antiderivative; above x-axis, increasing anti-derivative
Find the function s(t) satisfying ds/dt = -4 + 3cost and s(0) = 2
C = 2; s(t) = -4t + 3sint + 2
Trig identities
Cos^2x + sin^2x = 1; tan^2x + 1 = sec^2x
Evaluate: S5 -5 (x-√25-x²)dx
Draw and split; S5 -5 (x)dx - S5 -5 (√25-x²)dx; 1/2(5)(5) + 1/2(5)(5) - 25π/2 = -25π/2 u²
Use properties of integrals to determine the value of I = S5 0 f(x)dx when S7 0 f(x)dx = 9, S7 5 f(x)dx = 7
Draw on number line; I=2
Evaluate: S2 -1 |x|dx
Draw; 1/2(1)(1) + 1/2(2)(2) = 2.5u²
Evaluate: S10 0 |x-5|dx
Draw; 1/2(5)(5) + 1/2(5)(5) = 25u²
Evaluate: S2 -1 (1-x)dx
Draw; S2 -1 (1-x)dx; 1/2(2)(2) - 1/2(1)(1) = 2-1/2 = 1.5u²
Evaluate: S-2 -3 ((y⁶-7y³)/y⁵)dy
1/2y² + 2/y]-2 -3 = -3/2 - 3 = -11/3
Find the general indefinite integral: S(x² + 1 + (1/x²+1))dx
1/3x³ + x + tan⁻¹x + C
Find the general indefinite integral: S(y³ + 1.8y² - 2.4y)dy
1/4y⁴ + 3/5y³ - 6/5y² + C
n=
# of rectangles
Find F'(x): F(x) = Sx² 1 (t³ - 4t² + 2)dt
((x²)³ - 4(x²)² + 2)(2x) - (1³ - 4(1)² + 2) = 2x⁷ - 8x⁵ + 4x + 1
Find F'(x): F(x) = Sx -3 (-t² + 5)dt
(-(x)² + 5) - (-(-3)² + 5) = -x² + 9
Evaluate: Sπ/4 0 tan²xdx
(sec²x - 1) = tanx -x]π/4 0 = 3π/4
Evaluate the integral: S-2 -5 (-x³ - 10x² - 32x - 36)dx
-1/4x⁴ - 10/3x³ - 16x² -36x]-2 -5; [-1/4(-2)⁴ - 10/2(-2)³ - 16(-2)² - 36(-2)] - [-1/4(-5)⁴ - 10/3(-5)³ -16(-5)² - 36(-5)] = -39/4
Evaluate the integral: S1 0 (-x³ - 2x² - x + 3)dx
-1/4x⁴ - 2/3x³ - 1/2x² + 2x]1 0; [-1/4(1)⁴ - 2/3(1)³ - 1/2(1)² + 3(1)] - 0 = 19/12
Evaluate the integral: S4 1 (-x²/2 + 3x - 5/2)dx
-1/6x³ + 3/2x² - 5/2x]4 1; [-1/6(4)³ + 3/2(4)² -5/2(4)] - [-1/6(1)³ + 3/2(1)² -5/2(1)] = 9/2
Evaluate the integral: S2 1 (1/x² - 4/x³)dx
-1/x + 2/x²]2 1 = (-1/2 + 2/2²) - (-1/1 + 2/1²) = -1
Evaluate the integral: S4 1 ((√y-y)/y²)dy
-2/√y - ln|y|]4 1 = (-2/√4 - ln|4|) - (-2/√1 - ln|1|) = 1-ln(4) = 1-2ln2
Evaluate the integral: S3 0 (2insx-e^x)dx
-2cosx - e^x]3 0 = -2cos(3) + 3
Evaluate the integral: S-8x³dx
-2x⁴ + C
Find F'(x): F(x) = S3x 2 -4(t+1)^1/3 dt
-4(3x+1)^1/3(3) - (-4(2+1)^1/3) = -12(3x+1)^1/3 + 12√3
Evaluate the integral: S12/x⁴ dx
-4/x³ + C
Evaluate the integral: S5cscxcotx dx
-5cscx + C
Find the indefinite integral: S(-6-6tan²θ)dθ
-6S(1+tan²θ)dθ = -6tanθ + C
Sa b f(x)dx =
-Sb a f(x) dx
Find the derivative: y = S1 sinx (√1+t²)dt
-Ssinx 1 (√1+t²)dt = -cosx(√1+t²)
Find the derivative: G(x) = S1 x (cos√t)dt
-Sx 1 cos√tot = -cos√t
Sa x f(t)dt =
-Sx a f(t)dt
Find dy/dx; y = S0 ³√x cos(t³)dt
-cos(x) = -1/3x^-2/3cosx
Find the general indefinite integral: S(csc²t - 2e^t)dt
-cotx - 2e^t + C
Evaluate the integral: Sπ/3 π/4 csc²θdθ
-cotθ]π/3 π/4 = -cot(π/3) + cot(π/4) = -(√3/3) + 1 = 1-√3/3
Evaluate: S3π/4 π/4 cscθcotθ dθ
-cscθ]3π/4 π/4 = 0
Find d/dx Sx⁴ 0 e^-t dt
-e^-t]x⁵ 0 = -e^-x^5; 5x⁴e^-x^-5
Evaluate the integral: S1 -2 -e^xdx
-e^x]1 -2; -e + e⁻²
Evaluate the integral: S -secxtanx dx
-secx + C
Evaluate: S3 -5 |x|dx
-x: -1/2x² (-5->0) = 25/2; x: 1/2x² (0->5) = 9/2; 25/2 + 9/2 = 17
Evaluate: S√2 1 (u⁷/2 - 1/u⁵)du
1/16u⁸ + 1/4u⁻⁴]√2 1 = .75
Find the antiderivative: f(x) = 1/2x² - 2x + 6
1/2 x²+1/2+1 - 2x^1+1/1+1 + 6x^0+1/0+1 = 1/6x³ - x² + 6x + C
Find the derivative: h(x) = S√x 1 (z²/z⁴+1)dz
1/2x(z²/z⁴+1)
