INSTRUCTORS Raffi Hovasapian John Zhu

Start learning today, and be successful in your academic & professional career. Start Today!

• ## Related Books

 0 answersPost by Mohamed Al Mohannadi on February 2 at 06:44:19 AMExample 3 is wrong, you are getting the area from -0.77 to 1.42 under the graph of f(x) we dont want that area we want the area under that graph BUT ALSO bounded by y=2.I noticed from comments and videos there are mistakes in videos almost every videos. I rely for this for my AP Calculus BC exam which I need a 4 on Please be careful 0 answersPost by Abdelrahman Megahed on January 2, 2013EXAMPLE 2 YOU SAID f(y)=y^2 then you write it as g(y) not f(y) 0 answersPost by liqun zhu on August 3, 2012At 16:00, when you were integrating ln(x+2), I believe you thought you were differentiating it and used the derivative rule instead of the integral rule of natural log.[ âˆ« ln(x)dx = x âˆ™ (ln(x) - 1) ]

### Area Between Curves

• Find which function or curve is above the other
• Apply integral directly
• Evaluate for resulting area
• If not given bounds of integration:
• Find points of intersection
• Take integral from left most intersection to right most intersection

### Area Between Curves

Find the area enclosed by y = 4 and y = x2
• First, we find the bounds. To do so, we set the equations equal to each other
• x2 = 4
• x = ±2
• A = ∫−22 4 − x2  dx
• A = (4x − [(x3)/3]) |−22
• A = 8 − [8/3] − (−8 − [(−8)/3])
• A = 16 − [16/3]
A = [32/3]
Find the area under the velocity curve v(t) = 3t2 + t from t = 0 to t = 2
• A = ∫02 3t2 + t − 0  dt
• The subtraction by zero has no effect. It is just to illustrate that we're taking the x-axis as the lower-bound.
• A = (t3 + [(t2)/2]) |02
• A = 8 + 2 − 0 − 0
• The area under a velocity curve also happens to be the displacement!
• t1t2 v(t) dt = x(t2) − x(t1)
A = 10
Find the area under y = ex from x = 0 to x = ln10
• A = ∫0ln10 ex dx
• A = ex |0ln10
• A = eln10 − e0
• A = 10 − 1
• A = 9
A = 9
Find the area under y = e[t/3] from t = 0 to t = 6
• A = ∫06 e[t/3] dt
• u = [t/3]
• du = [1/3] dt
• A = 3 ∫06 eu du
• A = 3 e[t/3] |06
• A = 3 (e2 − e0)
A = 3(e2 − 1)
Find the area bounded by y = x2 and y = |x|
• Find the bounds of the integral
• x2 = |x|
• x2 = √{x2}
• x4 = x2
• x2 = 1
• x = ±1
• y = |x| has different behavior on either side of zero. We can treat this as a piecewise function with the area as the sum of two different integrals
• A = ∫−10 −x − x2  dx + ∫01 x − x2  dx
• A = (−[(x2)/2] − [(x3)/3]) |−10 + ([(x2)/2] − [(x3)/3]) |01
• A = (−0 − 0) − (−[((−1)2)/2] − [((−1)3)/3]) + ([1/2] − [1/3]) − 0
• A = [1/2] − [1/3] + [1/2] − [1/3]
• A = 1 − [2/3]
A = [1/3]
Find the area bound by x = y3 and the y-axis from y = 0 to y = 2
• A = ∫02 y3 dy
• A = [(y4)/4] |02
• A = [(24)/4] − 0
A = 4
Find the area bound by y = sinx and the x-axis from x = 0 to x = 2π
• A = ∫0 sinx  dx
• A = − cosx |0
• A = −cos2π− (− cos0)
• A = −1 + 1 = 0
• An area of zero doesn't quite make sense here, so using the straight integral is insufficient. What we can do is treat this as two separate integrals, one where the area is above the x-axis and one where it is below and add their effective area.
• sinx = 0, x = 0, π, 2π in this interval.
• A = ∫0π sinx  dx + (−∫π sinx  dx)
• The negative sign is there to compensate for the "negative area"
• A = −cosx |0π + cosx |π
• A = − cosπ+ cos0 + cos2π− cosπ
• A = 1 + 1 + 1 + 1
A = 4
Find the area bound by x = y2 and y = x − 2
• We could find the integral in terms of dx, but the we would have to split it up into at least two separate integrals. Besides, x = y2 is not a function of x (there are two outputs for every x input).
• x = y2
• x = y + 2
• Find the limits of the integral
• y2 = y + 2
• y2 − y − 2 = 0
• (y + 1)(y − 2) = 0
• y = 2, −1
• A = ∫−12 (y + 2) − y2 dy
• A = ([(y2)/2] + 2x − [(y3)/3]) |−12
• A = (2 + 4 − [8/3]) − ([1/2] − 2 + [1/3])
• A = 8 − [1/2] − [9/3]
• The area above the
A = [9/2]
Find the area bound by y = x2 + 1 and y = −x2 + 3
• Find the limits of the integral
• x2 + 1 = −x2 + 3
• 2x2 = 2
• x2 = 1
• x = ±1
• A = ∫−11 (−x2 + 3) − (x2 + 1)  dx
• A = ∫−11 −2x2 + 2  dx
• A = (−[(2x3)/3] + 2x) |−11
• A = −[2/3] + 2 − (−[(−2)/3] − 2)
• A = −[2/3] + 2 − [2/3] + 2
• A = 4 − [4/3]
A = [8/3]
Find the area bound by y = x2 and y = 2x
• Find the limits
• x2 = 2x
• x2 − 2x = 0
• x(x − 2) = 0
• x = 0, 2
• A = ∫02 2x − x2 dx
• A = x2 − [(x3)/3] |02
• A = 4 − [8/3] − 0
A = [4/3]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.