555 Video Lessons
Select the section below that fits the topic that you need help on. All our 555 Calculus videos are linked to our YouTube Channel and are Free to watch. Enjoy!
LIMITS – 60 Video Lessons
EVALUATING LIMITS
- lim 5 , x -> -1
- lim (-x + 2) , x -> -5/2
- lim (x^3-x^2-4) , x -> 2
- lim (-(x^2)/2 + 2x + 4) , x -> 1
- lim -√(x + 3) , x -> 3
- lim -√(2x + 4) , x -> 3/2
- lim – (x – 4)/(x^2 – 6x + 8) , x -> 1
- lim (-x – 3)/(x^2 + x + 1) , x -> 3/2
- lim sin(x) , x -> π
- lim 2cos(x) , x -> 3π/4
- Give an example of a limit that evaluates to 4
- Give an example of a limit of a quadratic function where the limit evaluates to 9
LIMITS AT JUMP DISCONTINUITIES AND KINKS
- lim x->-1^+ 4x+4/|x+1|
- lim x->-1^- f(x), f(x)={-x-8 when x≤-1; -x^2-4x-4 when x>-1
- lim x->-3 f(x), f(x)={-x^2-10x-24 when x≤-3; 2x+3 when x>-3
- lim x->-1 f(x), f(x)={x when x<-1; -x^2+2x when x≥-1
- lim x->-1^- f(x), f(x)={-x-3 when x≤-1; x+1 when x>-1
- lim x->-2 f(x), f(x)={-x^2-4x-5 when x≤-2; -1 when x>-2
- lim x->0^+ f(x), f(x)={1 when x≤0; -x^2+4x-3 when x<0
- lim x->0^- |x|/x
- lim x->0^+ floor(-2x+1)
- lim x->1 f(x), f(x)={x/2+9/2 when x<1;x^2-6x+10 when x≥1
- lim x->-1 3|x+1|/(x+1)
- lim x->-2 f(x), f(x)={x^2 when x≤-2; -x/2+3 when x>-2
- Give an example of a two-sided limit of a piecewise function where the limit does not exist.
- Give an example of a two-sided limit of a function with an absolute value where the limit does not exist.
LIMITS AT REMOVABLE DISCONTINUITIES
- lim x->2 f(x) where f(x)={-x^2+2 when x≠2; -5 when x=2
- lim x->-2 -(x^2-4)/(x+2)
- lim x->3 (x^2-7x+12)/(x-3)
- lim x->-3 (x+3)/(x^2+2x-3)
- lim x->0 f(x) where f(x)={x+1 when x≠0; 2 when x=0
- lim x->3 f(x) where f(x)={2+x/2 when x≠3; 2 when x=3
- lim x->1 -(x^2-1)/(x-1)
- lim x->5 -(x^2-5x)/(x-5)
- lim x->2 -(x^2-x-2)/(x-2)
- lim x->-5 (x^2+3x-10)/(x+5)
- lim x->0 ((1/(-4+x))+1/4)/x
- lim x->-3 x/((1/(3+x))-1/3)
- lim x->5 (x-5)/(√(x+4)-3)
- lim x->3 (√(x+6)-3)/(x-3)
- Give an example of a limit of a rational function where the limit at -1 exists, but the rational function is undefined at -1.
- Give two values of a where the limit cannot be solved using direct evaluation. Give one value of a where the limit can be solved using direct evaluation. lim x->a x/((1/(-2+x))+1/2)
LIMITS AT ESSENTIAL DISCONTINUITIES
- lim x->-3^+ (x+2)/(x^2+5x+6)
- lim x->-4 (x^2)/(4x+16)
- lim x->-2^+ 3x/(x+2)
- lim x->-1^+ (x^2)/(x+1)
- lim x->-3^- 2x/(x+3)
- lim x->-2^+ 1/(x^2-4)
- lim x->3^- -4x/(x-3)
- lim x->1 -3/(x-1)
- lim x->-2^- (x+2)/(x^2+x-2)
- lim x->-3^- -2/(x+3)
- lim x->π/4^- 2sec(2x)
- lim x->3π/4^+ 2tan(2x)
- Give an example of a right-sided limit that goes to ∞ as x goes to 5.
- Give an example of a left-sided limit that goes to ∞ as x goes to 5.
CONTINUITY – 17 Video Lessons
CONTINUITY
- f(x)={x^2+2x+1 when x<1; -x/2 when x≥1
- f(x)={1 when x≠5; 3 when x=5
- f(x)={2x-10 when x<2; 0 when x≥2
- f(x)=(x^2-x-2)/(x+1)
- f(x)=(x^2)/(2x+4)
- f(x)=-x/2-7/2 when x≤0; -x^2+2x-2 when x>0
- f(x)=-(x^2-x-12)/(x+3)
- f(x)=(x^2-x-6)/(x+2)
- f(x)=-(x^2)/(2x+4)
- f(x)=(x+1)/(x^2-x-2)
- f(x)=(x+1)/(x^2+x+1)
- f(x)=-(x^2)/(x-1)
- f(x)={x^2-4x+3 when x≠0; 3 when x=0
- f(x)={-x^2 when x≠1; 0 when x=1
- Give an example of a function with discontinuities at x=1, 2, and 3.
- Of the six basic trigonometric functions, which are continuous over all real numbers? Which are not? What types of discontinuities are there?
DIFFERENTIATION – 147 Video Lessons
AVERAGE RATES OF CHANGE
- y=x^2-x+1; [0,3]
- y=-1/(x-2); [-3,-2]
- y=-x^2+x+2; (-2,-4), (1,2)
- y=-1/x; (1,-1), (3,-1/3)
- y=x^2+2; [-2,-3/2]
- y=2x^2-2x+1; [-1,-1/2]
- y=-1/(x+2); [-1,-1/2]
- y=2x^2+x+2; [0,1/2]
- y=-x^2-2; (1,-3), (3/2,17/4)
- y=1/(x+3); (-1,1/2), (-1/2,2/5)
- y=1/(x-1); (-2,-1/3), (-3/2,-2/5)
- y=-1/x; (1,-1), (3/2,-2/3)
- The police have accused a driver of breaking the speed limit of 60 miles per hour. As proof, they provide two photographs. One photo shows the driver’s car passing a toll booth at exactly 6 PM. The second photo shows the driver’s car passing another toll booth 31 miles down the highway at exactly 6:30 PM. Does the photo evidence prove that the driver broke the speed limit during this time?
INSTANTANEOUS RATES OF CHANGE
POWER, CONSTANT, AND SUM RULES
- y = 5
- f(x) = 5x^18
- y = 4x^5 + x
- f(x) = 4x^4 – 5x – 3
- y = 3x^(5/4)
- y = (5/4)x^(2/3)
- y = -4x^-5
- y = 3/x^3
- y = x^(2/3)
- f(x) = -2 fourth root (x)
- y = (2/3)x^4 + 5x – x^-3
- y = (-1/2)x^4 + 3x^(5/3) + 2x
- y = -3r^5 – 5r^2
- f(s) = (-3/s^2) – (4/s^4)
- f(x) = (2/3)x^(3/2) – (3/4)x^(3/5)
- h(s) = sqrt(2)cbrt(s) + sqrt(2)5th-rt(s)
- y = 5c
- y = 4ax^3a – bx^3c
DIFFERENTIATION – PRODUCT RULE
- y = -x^3(3x^4 – 2)
- f(x) = x^2(-3x^2 – 2)
- y = (-2x^4 – 3)(-2x^2 + 1)
- f(x) = (2x^4 – 3)(x^2 + 1)
- f(x) = (5x^5 + 5)(-2x^5 – 3)
- f(x) = (-3 + x^-3)(-4x^3 + 3)
- y = (-2x^4 + 5x^2 + 4)(-3x^2 + 2)
- y = (x^4 + 3)(-4x^5 + 5x^4 + 5)
- y = (5x^4 – 3x^2 – 1)(-5x^2 + 3)
- y = (-10x^2 – 7(5th rt(x^2)) + 9)(2x^3 + 4)
- y = (5 + 3x^-2)(4x^5 + 6x^3 + 10)
- y = (-6x^4 + 2 + 6x^-4)(6x^4 + 7)
- f(x) = (-7x^4 + 10x^2/5 + 8)(x^2 + 10)
- Prove (f • g)’ ≠ f’ • g’ for any functions f and g
DIFFERENTIATION – QUOTIENT RULE
- f(x) = 2/(2x^4 – 5)
- f(x) = 2/(x^5 – 5)
- f(x) = 5/(4x^3 + 4)
- y = (4x^3 – 3x^2)/(4x^5 – 4)
- y = (3x^4 + 2)/(3x^3 – 2)
- y = (4x^5 + 2x^2)/(3x^4 + 5)
- y = (4x^5 + x^2 + 4)/(5x^2 – 2)
- y = (3x^4 + 5x^3 – 5)/(2x^4 – 4)
- y = (x^3 – x^2 – 3)/(x^5 + 3)
- y = (x^4 + 6)/(3 – 4x^-4)
- y = (4x^4 – 4x^2 + 5)/(2x^5/3 + 3)
- Prove (f/g)’ ≠ f’/g’ for any functions f and g
DIFFERENTIATION – CHAIN RULE
- y = (x^3 + 3)^5
- y = (-3x^5 + 1)^3
- y = (-5x^3 – 3)^3
- y = (5x^2 + 3)^4
- f(x) = 4th-rt(-3x^4 – 2)
- f(x) = sqrt(-2x^2 + 1)
- f(x) = cbrt(-2x^4 + 5)
- y = (-x^4 – 3)^-2
- y = (3x^3 + 1)(-4x^2 – 3)^4
- y = ((x^3 + 4)^5)/(3x^4 – 2)
- y = ((x + 5)^5 – 1)^4
- y = (5x^3 – 3)^5((4th-rt(-4x^5 – 3))
- Give a function that requires three applications of the chain rule and differentiate
DIFFERENTIATION – NATURAL LOGS AND EXPONENTIALS
APPLICATIONS OF DIFFERENTIATION – 136 Video Lessons
ROLLE’S THEOREM
- y=x^2+4x+5; [-3,-1]
- y=x^3-2x^2-x-1; [-1,2]
- y=-x^3+2x^2+x-6; [-1,2]
- y=x^3-4x^2-x+7; [-1,4]
- y=-x^3+2x^2+x-1; [-1,2]
- y=x^3-x^2-4x+3; [-2,2]
- y=(-x^2-2x+15)/(-x+4); [-5,3]
- y=(x^2-2x-15)/(-x+6); [-3,5]
- y=(-x^2+2x+15)/(x+4); [-3,5]
- y=(x^2+x-6)/(-x+3); [-3,2]
- y=-2sin(2x); [-π,π]
- y=sin(2x); [-π,π]
- y=(x^2-x-12)/(x+4); [-3,4]
- y=(-x^2-2x+8)/(-x+3); [-4,2]
- y=(-x^2+36)/(x+7); [-6,6]
- y=(-x^2+4)/(4x); [-2,2]
- y=2tan(x); [-π,π]
- y=-2cos(2x); [-π,π]
MEAN VALUE THEOREM
- y=-x^2+8x-17; [3,6]
- y=x^3-9x^2+24x-18; [2,4]
- y=-(x^2/2)+x-1/2; [-2,1]
- y=(x^2/2)-2x-1; [-1,1]
- y=x^3+3x^2-2; [-2,0]
- y=-x^3+4x^2-3; [0,4]
- y=(x^2-9)/3x; [1,4]
- y=x^2/(2x-4); [-4,1]
- y=-(-2x+6)^(1/2); [-2,3]
- y=-(-5x+25)^(1/2); [3,5]
- y=-x^2/(4x+8); [-3,-1]
- y=(-x^2+9)/4x; [1,3]
- y=-(6x+24)^(2/3); [-4,-1]
- y=(x-3)^(2/3); [1,4]
- Use the Mean Value Theorem to prove that |sin(a)-sin(b)| ≤ |a-b| for all real values of a and b where a≠b.
INTERVALS OF INCREASE AND DECREASE
- y=-x^3+2x^2+2
- y=x^3-11x^2+39x-47
- y=-x^4+3x^2-3
- y=x^2/(4x+4)
- y=(3x^2-3)/x^3
- y=(2x-8)^(2/3)
- y=-(1/5)(x-4)^(5/3)-2(x-4)^(2/3)-1
- If functions f and g are increasing on an interval, show that f+g is increasing on the same interval.
- Give an example where functions f and g are increasing on the interval (-∞,∞), but where f–g is decreasing.
RELATIVE EXTREMA
- y=x^3-5x^2+7x-5
- y=x^3-6x^2+9x+1
- y=-x^3-3x^2-1
- y=x^4-2x^2+3
- y=x^4-x^2
- y=-2/(x^2-4)
- y=(2x-8)^(2/3)
- y=-(1/5)(x-4)^(5/3)-2(x-4)^(2/3)
- Give an example function f(x) where f”(0)=0 and there is no relative minimum or maximum at x=0.
- Give an example function f(x) where f”(0)=0 and there is no relative maximum at x=0.
RELATED RATES
- Water leaking onto a floor forms a circular pool. The radius of the pool increases at a rate of 4 cm/min. How fast is the area of the pool increasing when the radius is 5 cm?
- Oil spilling from a ruptured tanker spreads in a circle on the surface of the ocean. The area of the spill increases at a rate of 9π m^2/min. How fast is the radius of the spill increasing when the radius is 10 m?
- A conical paper cup is 10 cm tall with a radius of 10 cm. The cup is being filled with water so that the water level rises at a rate of 2 cm/sec. At what rate is water being poured into the cup when the water level is 8 cm?
- A spherical balloon is inflated so that its radius (r) increases at a rate of 2/r cm/sec. How fast is the volume of the balloon increasing when the radius is 4 cm?
- A 7 ft tall person is walking away from a 20 ft tall lamppost at a rate of 5 ft/sec. Assume the scenario can be modeled with right triangles. At what rate is the length of the person’s shadow changing when the person is 16 ft from the lamppost?
- An observer stands 700 ft away from a launch pad to observe a rocket launch. The rocket blasts off and maintains a velocity of 900 ft/sec. Assume the scenario can be modeled as a right triangle. How fast is the observer to rocket distance changing when the rocket is 2400 ft from the ground?
INDEFINITE INTEGRATION – 99 Video Lessons
INTEGRATION TRIG FUNCTIONS
INTEGRATION INVERSE TRIGONOMETRIC FUNCTIONS
INTEGRATION BY SUBSTITUTION (POWER RULE)
- ∫-15x^4(-3x^5-1)^5 dx; u=-3x^5-1
- ∫-16x^3(-4x^4-1)^-5 dx; u=-4x^4-1
- ∫-(8x^3)/((-2x^4+5)^5) dx; u=-2x^4+5
- ∫(5x^4+5)^(2/3)•20x^3 dx; u=5x^4+5
- ∫((5+ln(x))^5)/x dx; u=5+ln(x)
- ∫4sec(4x)•tan(4x)•sec^4(4x) dx; u=sec(4x)
- ∫36x^3(3x^4+3)^5 dx; u=3x^4+3
- ∫x(4x-1)^4 dx; u=4x-1
- ∫-9x^2(-3x^3+1)^3 dx
- ∫12x^3(3x^4+4)^4 dx
- ∫-12x^2(-4x^3+2)^-3 dx
- ∫(3x^5-3)^(3/5)•15x^4 dx
- ∫(-2x^4-4)^4•-32x^3 dx
- ∫(e^4x-4)^(1/5)•8e^4x dx
- ∫x(4x+5)^3 dx
- ∫5x√(2x+3) dx
INTEGRATION BY SUBSTITUTION (LOGARITHMS AND EXPONENTIALS)
DEFINITE INTEGRATION – 75 Video Lessons
FIRST FUNDAMENTAL THEOREM OF CALCULUS
- ∫(-x^3+3x^2+1) dx from -1 to 3
- ∫(x^4+x^3-4x^2+6) dx from -2 to 1
- ∫(2x^2-12x+13) dx from 1 to 3
- ∫(-x^3+3x^2-2) dx from 0 to 3
- ∫(x^5-4x^3+4x+4) dx from -1 to 0
- ∫4x^(1/3) dx from -3 to 0
- ∫-(4/x^3) dx from -4 to -1
- ∫4/x dx from -3 to -1
- ∫2cos(x) dx from (-π/4) to (-π/6)
- ∫1/(x√(x^2-1) dx from √2 to 2
- ∫5(2x+4)^(1/3) dx from -3 to -2
- ∫2/(2x+4)^3 dx from -1 to 2
- ∫e^(2x-2) dx from -1 to 1
- ∫(-x+|-3x-9|) dx from -4 to -2
- ∫f(x) dx from 0 to 3 where f(x)={x/2-1 when x≤2, x^2-6x+8 when x>2
- ∫-|x^2+4x| dx from -5 to 1
SUBSTITUTION FOR DEFINITE INTEGRALS
- from -1 to 0 ∫8x/(4x^2+1)^2 dx; u=4x^2+1
- from 0 to 1 ∫-12x^2(4x^3-1)^3 dx; u=4x^3-1
- from -1 to 2 ∫6x(x^2-1)^2 dx; u=x^2-1
- from 0 to 1 ∫24x/(4x^2+4)^2 dx; u=4x^2+4
- from -3 to 0 ∫-8x/(2x^2+3)^2 dx; u=2x^2+3
- from 0 to 1 ∫16x/(4x^2+4)^2 dx; u=4x^2+4
- from -1 to 0 ∫18x^2(3x^3+3)^2 dx; u=3x^3+3
- from 0 to 1 ∫-8x/(4x^2+2)^2 dx; u=4x^2+2
MEAN VALUE THEOREM FOR INTEGRALS
- f(x)=-x^2-2x+5; [-4,0]
- f(x)=-x^4+2x^2+4; [-2,1]
- f(x)=-x^2/2+x+3/2; [-3,1]
- f(x)=4/x^2; [-4,-2]
- f(x)=-x^3+7x^2-11x+3; [1,5]
- f(x)=-x^5+3x^3; [0,1]
- f(x)=4x^(1/2); [0,3]
- f(x)=x^5-2x^3+x; [-1,0]
- f(x)=1/x; [2,3]
- f(x)=x^5-4x^3+2x-1; [-2,2]
- f(x)=-x^5+4x^3-5x-3; [-2,0]
- f(x)=x^5-2x^3-2; [-1,1]
- f(x)=-x+2; [-2,2]
- f(x)=-x^2-8x-17; [-6,-3]
- f(x)=-3(2x-6)^(1/2); [3,5]
- f(x)=4/(2x+6)^2; [-6,-5]