AP Calculus BC Unit 6 Integration and Accumulation of Change

Unit 6: Integration and Accumulation of Change ● The area between a function on a graph and the x-axis can give us the accumulation of change. ○ For example, finding the area between a function that gives an object’s velocity with respect to it’s time and the x-axis would give us the total change in position. ○ Sometimes, you can just find the area by using basic geometry, if the area of the function is a square, a rectangle, etc. ■ Keep in mind that if the graph of the function is below the x-axis, then the area would be negative as well. If the graph is above the x-axis, then the area would be positive. ○ However, most of the time, the graph of the function is not an ideal shape, in which basic geometry cannot be used to find the area of the curve, such as the graph below: ○ The units of accumulation of change are found by multiplying the units of the independent variable to the units of the dependent variable. ● Integrals: The integral of a function is basically the ANTIDERIVATIVE of the given function. ○ The symbol to denote that you need to find the integral is: ∫ ○ Basically, if you are asked to find ∫2x dx, then you’re trying to find a function that when you take the derivative of it, you get 2x. ■ “Dx” just represents the change in x, since you normally find the area by multiplying the width and the length (in simple terms). The width of the graph is the change in x, or “dx”. The “length” of the graph is typically going to be the equation of the function, which in this case, is 2x (this will be explained more in depth later). ● Riemann Sums (RS): an approximation of an integral that’s used when the function isn’t given ○ The approximation is made by adding the areas of a certain number of rectangles within the bounded graph (look at the picture below) ■ Look at the picture above. The areas of each graph can be approximated by adding up the areas of all the rectangles that are drawn into the graph. The height of the rectangles can vary depending on the type of Riemann Sum asked for. ● In addition, notice how the greater the number of rectangles there are, the more accurate the approximation of the area of the function bound between the function curve and the x-axis is (you know this because you see less white space showing in the graph with the yellow rectangles, than the one with blue rectangles. ○ Riemann Sums are usually asked for when the problem doesn’t give you the equation of the function, and instead just gives you values of the corresponding f(x) values to certain x values (typically give you a table) 8 ■ Ex. Find∫2 f(x)dx using a [right/midpoint/left/trapezoidal] Riemann sum with 3 subintervals. x 2 4 6 8 f(x) 3 16 21 12 ● The problem would specify the type of Riemann sum they’re looking for and give you a table of values for you to use. ○ There are 4 types: left, right, midpoint, and trapezoidal. ■ The trapezoidal Riemann Sum is the most accurate, then the midpoint RS, and then the left and right Riemann sums are the least accurate. ■ Right RS, Left RS, and Midpoint RS: ● These three RS are approached similarly; the only difference would be the values that are added (the ones in the brackets) 𝑏 ● *remember that integrals are written in the form of ∫𝑏 f(x)dx. Thus, in the equation above, ‘b’ represents the upper limit of the integral, and ‘a’ represents the lower limit. ○ In the example above, b=8 and a=2. ● ‘N’ represents the number of subintervals that the questions asks you to divide the curve up into. Remember from the picture about Riemann sums that you can break up the curve into a lot of subintervals (so a lot of rectangular areas to add up), or even just a couple of subintervals only. ○ In the example above, n=3 ● After recognizing the amount of subintervals wanted, break the table up into the correct amount of intervals. ○ In the example above, because the question asks for 3 subintervals, it’d only make sense to break the table up into intervals from x=2 to x=4, x=4 to x=6, and x=6 to x=8 ■ Make sure that all the intervals are evenly spaced! ● The picture above explains which values to use for the right and left RS. ● For midpoint RS, the approach is, again, very similar to that of right and left RS, but the values used are different. ○ Midpoint means “middle”. Therefore, the values that are added up to find midpoint RS are the middle values in each interval. ■ In the example above, x=3 is the midpoint of the interval between x=2 and x=4. To find f(3), you just average the values of f(2) and f(4), so f(3) = 19/2. Then add f(3) to f(5) and f(7). ● Note that the formulas for RS above only work if the intervals are EVEN. In this case, each interval had a width of 2. x 2 3 7 8 f(x) 3 19 25 12 ● However, there are cases when they give you a table such as this one, where the intervals don’t have the same width. ○ In this case, you’d still divide the table up into the 3 intervals (2,3), (3,7), (7,8), except the intervals aren’t evenly spaced anymore. ○ Now, you’d have to subtract the endpoints of each interval to account for the different widths. ■ (1/3) [(3-2)(19) + (7-3)(25) + (8-7)(12) → this is for RIGHT RS ● Notice how instead of multiplying all the added values to (b-a)/n, you have to multiply each individual value by the width of the interval. ■ Trapezoidal RS: ● *Take note that the formula above only works if the widths of the trapezoids are equal. However, if a question gives you a table and when you have divided up the values into the required number of “sections” ■ Approximation of Riemann Sums ● If a function is concave up, and a right RS is used, then the answer you get is an overapproximation of the area under the curve. If a left RS is used to find the approximation of the area under the curve of a concave up function, then it’d be an underapproximation. ● If a function is concave down, then the right RS would be an underapproximation, and the left RS would be an overapproximation. ○ The definite integral of a continuous function over a closed interval [a,b] is equal to the limit of Riemann sums as the width of the subintervals approaches 0 (indicated by the formula below) ● Power rule for Antiderivatives ○ EXCEPTION: when n= -1 ■ *Remember that the derivative of f(x) = ln(x) is f’(x) = 1/x ● Thus, ∫1/x dx = ln(x) + C ○ Ex. ∫𝑥 4 dx = 𝑥 4+1 / (4+1) + C = 𝑥 5 /5+ C ■ And if you take the derivative of 𝑥 5 /5 + C, you’d get 𝑥 4 ■ *’C’ refers to any constant. Remember that the derivative of a constant is 0, so when taking the antiderivative of a function, you need to add ‘C’ to what you get. ● The Fundamental Theorems of Calculus 4 ■ Ex. ∫1 𝑥 4 dx = F(4) - F(1) ● 1st, find the antiderivative of the function, which as we saw above, equals 𝑥 5 /5 + C. Because this is a DEFINITE integral (bounded by x values), the “c” isn’t necessary. ● Now that you know that F(x) = 𝑥 5 /5, solve for F(4) - F(1), which equals 45 /5- ⅕ = 1023/5 𝑥 𝑡 4 dt, find g’(x) at x = 2. ■ Ex. if g(x) = ∫1 ● According to the Fundamental Theorem of calculus, 𝑥 g’(x) = (d/dx) ∫1 𝑡 4 dt = 𝑥 4 . Then just plug x=2 into your equation, and you will get g’(2) = 16 ● Graphs and definite integrals: ○ Let’s say that g(x) = f’(x). Find f(35) if f(0) = 20. 35 ■ In order to find f(35), you need to find the integral ∫0 𝑓′(𝑥)dx, which according to the Fundamental Theorem of Calculus, should equal to f(35)35 f(0) = ∫0 𝑓′(𝑥)dx. Isolate f(35), and you’d get 35 f(35) = f(0) + ∫0 35 𝑓′(𝑥)dx = 20 + ∫0 𝑓′(𝑥)dx. ● Since we don’t know what the equation of the function is, you can’t use power rule to antidifferentiate and solve for 35 ∫0 𝑓′(𝑥)dx. However, we’re given the graph of f’(x). Remember that finding the integral of a function can be done by finding the area between the curve, and the x-axis. 35 ○ Thus, ∫0 𝑓′(𝑥)dx = the area of the triangle from (0,10), the semicircle from (10,30), and the mini triangle from (30,35). 35 ■ ∫0 𝑓′(𝑥)dx = 100 - 50ℼ + 25/2 ■ f(35) = 20 + 100 - 50ℼ + 25/2 = 245/2 - 50ℼ ● Properties of Definite Integrals: 2 2𝑥 2 dx ○ Constants → ex. ∫0 ■ When asked to find the integral of a constant times a function, the integral 2 𝑥 2 dx , with the constant outside of the can be reformatted to 2∫0 integral, and solve the integral separately, before multiplying 2 to it. 2 2𝑥 2 − 𝑥dx ○ Addition and subtraction → ex. ∫0 2 ■ ∫0 2 2𝑥 2 − 𝑥dx is equal to ∫0 2 2𝑥 2 𝑑𝑥 − ∫0 𝑥dx ● You can integrate each function independently, and then add/subtract as needed ○ Reversal of limits of integration ■ Typically when you see an integral, the lower limit is less than the higher 2 𝑓(𝑥)→ 0 < 2, and so 0 is placed at the bottom. limit. Ex. ∫0 ● However, sometimes, an integral can be written like this: 0 ∫2 𝑓(𝑥)𝑑𝑥. In this case, you can reverse the limits, so that it’s written with the lower number on the bottom, but when you reverse them, you need to negate the function. 0 ○ Ex. if ∫2 2 f(x) dx = 5, then ∫0 f(x)dx = -5 ● Integrals of Trig Functions: ○ Again, it’s easiest to memorize the formulas for the integral of trig functions. ● U-Substitution: ○ Use u-substitution to reverse a function that was differentiated by chain rule. ■ This is an overview of how u-substitution works. The example below will show you how to implement u-substitution ■ Another way to solve this is to plug ‘𝑥 2 - 9’ back into ‘u’ after integrating ‘𝑢1/2 du’ and keep the interval from x=3 to x=5. ● Basically, the last step can look like this: ○ I = [(2/3)(𝑥 2 − 9)3/2 ]₃⁵ ℼ/2 ○ Ex. 2: ∫0 cosx(sinx)²dx ■ In this case, u=sinx (usually, the function that has the higher exponent is the function you use for ‘u’) ● Du = cos(x) dx, and you can rearrange this equation to get dx = du /cosx ■ Then, plug in your newly derived equation in dx, so that you get ℼ/2 ∫0 (cosx)u²(du/cosx) ℼ/2 ● Now you can cancel out the ‘cosx’, and are left with ∫0 u²du. ■ Integrate, and you’d get [𝑢³/3]₀ℼ/2 ● *note that 0 and ℼ/2 are the limits of the integral when the function was in terms of ‘x’, not ‘u’. ○ To account for this, you can change the limits by plugging x=0 and x=ℼ/2 into u=sinx, to find the limits for ‘u’ ■ When x=0, u=0 and when x=ℼ/2, u=1 ■ Now, replace the limits, and solve for [𝑢³/3]₀1 ○ The second way to account for this is to substitute the original function back into ‘u’ ■ You’d get: [(𝑠𝑖𝑛³𝑥)/3]₀ℼ/2 . Now you can solve for the answer. ● Integrating Functions using long division: ○ You can use long division to simplify the function before integrating to make it easier. ○ Ex. ∫ 𝑥2 +𝑥+1 dx. 𝑥2 +1 ■ Use long division to simplify: ∫ 𝑥 + 𝑥2 +1 1dx, and now it’s easier to integrate the function. You can use the additive rule to break up the integral into v∫ 𝑥 𝑑𝑥+ 𝑥2 +1 ∫ 1dx, and integrate the first integral using u-substitution and the second integral using power rule (or remember that the integral of any constant is that constant times x, so in this case 1(x) = x) ● Integration by Parts: ○ This is used to find the antiderivative of a function that was derived using the product rule. ○ Formula: ○ Ex. (from the Princeton 2020 AP Calculus BC book): find ∫ xsinx dx ■ 1) choose what function you want to be ‘u’ and ‘dv’ ● Technically, you can choose any function to be ‘u’ and ‘dv’, but there are cases where it’d be a lot easier to assign a certain function to ‘u’ ○ LIATE rule: use this to determine what function should be ‘u’ → the function that comes earlier in the acronym “LIATE” should be the one assigned as ‘u’ ■ L : logarithmic functions (ex. log(x), ln(x)) ■ I : inverse trig functions (ex. sin⁻¹(x), cos⁻¹(x)) ■ A : algebraic functions (ex. x, 𝑥 2 , 𝑥 2 + 1, 2x + 3) ■ T : trig functions (ex. sinx, cosx, tanx) ■ E : exponential functions (ex. 2𝑥 , 3𝑥 ) ● In this ex., breaking up the integral’s functions will give you x (algebraic function) and sinx (trig function) ○ Because ‘A’ comes before ‘T’ in the acronym “LIATE”, u = x, which means that dv = sinx. ■ 2) draw a table with the values of u,du,v,dv on it ● *this is optional, but it tends to help a lot, especially if you’re having a hard time understanding this ● In this ex., the table would look like this: u=x v = -cosx du = 1 dx dv = sinx dx ● *note that because sinx = dv, you need to integrate that function to find the function for ‘v’ ■ 3) plug the functions in the table into the formula ● ∫ xsinx dx = x(-cosx) - ∫ -cosx dx ■ 4) integrate the last part of the equation, and solve for the integral. If it’s a definite integral, follow the steps of normal integration to get a numerical answer. ● ∫ xsinx dx = -xcosx + sinx + C ● Integration using partial fractions 1: ○ Used to solve for integrals that contain rational expressions. ■ Ex. of rational expression: ∫ 3𝑥 + 5 dx (𝑥−2)(𝑥−1) ○ It’s important to first note that rational expressions can be split up into separate fractions like . ■ The picture above walks through the steps on how to split up the rational expression into 2 fractions that are added to each other. ● 1) set the rational expression equal to as shown above ○ (x-a) and (x-b) correlates to the denominator of the rational expression in the integral. ■ (x+1) is placed before (x-2) in the integral, so ‘a’ in ‘x-a’ = 1, and ‘b’ in ‘x-b’ = 2 ● 2) multiply both the numerator and denominator by the denominator of the rational expression that’s being asked to be integrated (in the ex. Above, it’s (x+1) and (x-2)) ○ This should leave you with 4x + 1 = A(x-2) + B(x+1) ● 3) You can reorganize the equation by distributing → 4x + 1 = Ax - 2A + Bx + B → 4x + 1 = x(A+B) + (-2A + B) ○ This makes it easier to solve for A and B ● 4) Solve for A and B using elimination ○ In this example, A + B = 4 and -2A + B = 1 ○ Using elimination, you’d eventually get A = 1 and B = 3 ● 5) Plug the values of A and B into the partial fraction and replace the rational expression in the integral with this new 1 partial fraction → ∫ 3 + dx (𝑥+1) (𝑥−2) ● 6) Now you can easily integrate the function using the additive rule → ∫ 1 dx (𝑥+1) 3 dx (𝑥−2) +∫ = ln(x+1) + 3ln(x-2) ● Integration using partial fractions 2: ○ Used if the denominator has a repeated function ■ Ex. from Princeton 2020 AP Calculus BC prep book: ∫ 2𝑥 + 4 (𝑥−1)2 ● 1) Rather than setting the rational expression equal to dx , instead, we set it equal to 2𝑥 + 4 𝐴 ○ In this ex. (𝑥−1)2 = (𝑥−1) + 𝐵 (𝑥−1)2 ● 2) solve for A and B using the same steps as before. ○ Multiply the whole equation by (𝑥 − 1)2 → 2x+4 = Ax + (B - A) ○ A = 2 because that’s the only value of A that will make the equation equal. Because (B-A) = 4, B = 6 ● 3) Plug in the values of A and B into the partial fraction equation, and then replace the rational expression with the new partial fraction equation. ● 4) Integrate using the additive rule ● Integration using partial fractions 3: 𝑝𝑥 2 + 𝑞𝑥 + 𝑟 ○ Use if the rational expression is in the form : (𝑥−𝑎)(𝑥 2 +𝑏) 2𝑥+1 ■ Ex. ∫(𝑥−2)(𝑥 2+1)dx ● Now, you’d set this rational expression equal to ○ 2𝑥+1 𝐴 𝐵𝑥+𝐶 = (𝑥−2) + (𝑥 2+1) (𝑥−2)(𝑥 2 +1) ■ *note that ‘Bx+C’ is the numerator for the fraction with the denominator of a quadratic equation, which is (𝑥 2 + 1)in this example. ● Solve for A, B, and C using the same method as before. ● At the end, when replacing A, B, C into the partial fraction equation, make sure you plug in C too! ● Integrate using the additive rule. You may have to break up the 𝐵𝑥+𝐶 fraction (𝑥 2 +1)even further to integrate more easily. ● Improper integrals: ○ An integral is improper if one/both of the limits of the integral is infinite, or if the integrand becomes infinite at 1+ points in the interval of integration ∞ ■ Ex. ∫0 𝑑𝑥 𝜋/2 (one of the limits is infinite) or ∫0 (𝑥 2 +1) ∞ ● Ex. 1 : ∫0 𝑑𝑥 𝜋 tanx dx ( tan(2)= ∞) (example taken from Princeton book) (𝑥 2 +1) ○ replace the upper limit with a, and take the limit of the 𝑎 integral as it approaches infinity → 𝑙𝑖𝑚 ∫0 𝑎→∞ 𝑑𝑥 (𝑥 2 +1) ○ integrate the function → 𝑡𝑎𝑛−1 (𝑎) − 𝑡𝑎𝑛−1 (0)= 𝑡𝑎𝑛−1 (𝑎) ○ Take the limit of the integrated function → 𝑙𝑖𝑚 𝑡𝑎𝑛−1 (𝑎) = 𝑎→∞ 𝑏/2 ● Ex. 2 : ∫0 𝜋 2 tanx dx ○ Similar to ex.1, replace the upper limit with a, and take the limit of the integral as it approaches infinity → 𝑎 𝑙𝑖𝑚𝜋 ∫0 𝑎→ tanx dx 2 ○ Integrate the function → ln|𝑠𝑒𝑐(𝑎)| − ln|𝑠𝑒𝑐(0)| ○ Take the limit of the integrated function → 𝑙𝑖𝑚𝜋 ln|𝑠𝑒𝑐(𝑎)| − ln|𝑠𝑒𝑐(0)|= ∞ − 0= undefined 𝑎→ 2