The topic with functions that we need to deal with is combining functions. We just computed the most general anti-derivative in the first part so we can use that if we want to. Now, we have a normal single integral so lets finish the integral by computing this. Lets work a couple of examples that involve other functions. First, determine where the quantity inside the absolute value bars is negative and where it is positive. Lets start with inverse sine. In the first integral we are differentiating with respect to \(y\) and we know that any function involving only \(x\)s will differentiate to zero and so when integrating with respect to \(y\) we need to acknowledge that there may have been a function of only \(x\)s in the function and so the constant of integration is a function of \(x\). Applications included are determining absolute and relative minimum and maximum function values (both with and without constraints), sketching the graph of a function without using a computational aid, determining the Linear Approximation of a function, LHospitals Rule (allowing us to So, we are really asking what angle \(y\) solves the following equation. Recall as well that two functions are inverses if \(f\left( {g\left( x \right)} \right) = x\) and \(g\left( {f\left( x \right)} \right) = x\). The boundary of \({D_{_1}}\) is \({C_1} \cup {C_3}\) while the boundary of \({D_2}\) is \({C_2} \cup \left( { - {C_3}} \right)\) and notice that both of these boundaries are positively oriented. LHospitals Rule and Indeterminate Forms In this section we will revisit indeterminate forms and limits and take a look at LHospitals Rule. While it might not seem like a useful thing to do with when we have the function there really are reasons that one might want to do this. Now lets see what would happen if we had integrated with respect to \(x\) first. Business Applications In this section we will give a cursory discussion of some basic applications of derivatives to the business field. Remember that when integrating with respect to \(y\) all \(x\)s are treated as constants and so as far as the inner integral is concerned the 2\(x\) is a constant and we know that when we integrate constants with respect to \(y\) we just tack on a \(y\) and so we get \(2xy\) from the first term. In other words, lets assume that \[{Q_x} - {P_y} = 1\] In this case the second term will have division by zero at \(y = 0\) and since \(y = 0\) is in the interval of integration, i.e. Linear Approximations In this section we discuss using the derivative to compute a linear approximation to a function. From a unit circle we can quickly see that \(y = \frac{\pi }{6}\). California voters have now received their mail ballots, and the November 8 general election has entered its final stage. On each of these intervals the function is continuous. Also, in this case there are no restrictions on \(x\) because tangent can take on all possible values. The fact that the first two terms can be integrated doesnt matter. So, doing the integration gives. In order to derive the derivatives of inverse trig functions well need the formula from the last section relating the derivatives of inverse functions. Recall from the indefinite integral sections that its easy to mess up the signs when integrating sine and cosine. In the previous section we gave the definition of the double integral. However, one of the more important uses of differentials will come in the next chapter and unfortunately we will not be able to discuss it until then. For this integral notice that \(x = 1\) is not in the interval of integration and so that is something that well not need to worry about in this part. If youre not sure of that sketch out a unit circle and youll see that that range of angles (the \(y\)s) will cover all possible values of sine. This was also a requirement in the definition of the definite integral. Given curves/regions such as this we have the following theorem. In this section we want to look at inequalities that contain absolute values. Next lets address the fact that we can use any anti-derivative of \(f\left( x \right)\) in the evaluation. Graph the first inequality and using the (0, 0) measure, test to see which side of the coordinate plane should be shaded. Well start with the definition of the inverse tangent. So, lets see how we can deal with those kinds of regions. From a unit circle we can see that \(y = \frac{\pi }{4}\). Solve each of the following inequalities. For problems 8 12 determine where the given function is discontinuous. Recall that when we talk about an anti-derivative for a function we are really talking about the indefinite integral for the function. As noted above we simply cant integrate functions that arent continuous in the interval of integration. Not much to do with this one other than differentiate each term. What this means for us is that when we do the integral all we need to do is plug in the first function into the integral. In this part \(x = 1\) is between the limits of integration. Lets first address the problem of the function not being continuous at \(x = 1\). We will compute the double integral by first computing. Contained in this site are the notes (free and downloadable) that I use to teach Algebra, Calculus (I, II and III) as well as Differential Equations at Lamar University. We will close out this section with an interesting application of Greens Theorem. Doing this gives. To see the proof of this see the Proof of Various Integral Properties section of the Extras chapter. \(\displaystyle f\left( x \right) = \frac{{4x + 5}}{{9 - 3x}}\), \(\displaystyle g\left( z \right) = \frac{6}{{{z^2} - 3z - 10}}\), \(g\left( x \right) = \left\{ {\begin{array}{rl}{2x}&{x < 6}\\{x - 1}&{x \ge 6}\end{array}} \right.\), \(h\left( t \right) = \left\{ {\begin{array}{rl}{{t^2}}&{t < - 2}\\{t + 6}&{t \ge - 2}\end{array}} \right.\), \(g\left( x \right) = \left\{ {\begin{array}{rc}{1 - 3x}&{x < - 6}\\7&{x = - 6}\\{{x^3}}&{ - 6 < x < 1}\\1&{x = 1}\\{2 - x}&{x > 1}\end{array}} \right.\), \(\displaystyle f\left( x \right) = \frac{{{x^2} - 9}}{{3{x^2} + 2x - 8}}\), \(\displaystyle R\left( t \right) = \frac{{8t}}{{{t^2} - 9t - 1}}\), \(\displaystyle h\left( z \right) = \frac{1}{{2 - 4\cos \left( {3z} \right)}}\), \(\displaystyle y\left( x \right) = \frac{x}{{7 - {{\bf{e}}^{2x + 3}}}}\), \(g\left( x \right) = \tan \left( {2x} \right)\), \(25 - 8{x^2} - {x^3} = 0\) on \(\left[ { - 2,4} \right]\), \({w^2} - 4\ln \left( {5w + 2} \right) = 0\) on \(\left[ {0,4} \right]\), \(4t + 10{{\bf{e}}^t} - {{\bf{e}}^{2t}} = 0\) on \(\left[ {1,3} \right]\). Well go through inverse sine, inverse cosine and inverse tangent in detail here and leave the other three to you to derive if youd like to. This is, You appear to be on a device with a "narrow" screen width (, \[A = \oint\limits_{C}{{x\,dy}} = - \,\oint\limits_{C}{{y\,dx}} = \frac{1}{2}\oint\limits_{C}{{x\,dy - y\,dx}}\], Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities. The latest Lifestyle | Daily Life news, tips, opinion and advice from The Sydney Morning Herald covering life and relationships, beauty, fashion, health & wellbeing Here is the definition for the inverse cosine. Note that in order to use these facts the limit of integration must be the same number, but opposite signs! The next topic of this section is a quick fact that can be used to make some iterated integrals somewhat easier to compute on occasion. Lets take a final look at the following integral. Section 4-2 : Iterated Integrals. So. Here are the steps for solving inequalities:. There is some alternate notation that is used on occasion to denote the inverse trig functions. This idea will help us in dealing with regions that have holes in them. You appear to be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities. This is here only to make sure that we understand the difference between an indefinite and a definite integral. Lets think of this double integral as the result of using Greens Theorem. If you think about it this was just a lot of work and all we got out of it was the result from Greens Theorem which we already knew to be true. Before working some examples there are some alternate notations that we need to acknowledge. There really isnt much to do here other than using the formula from above as noted above. Solution 1In this case we will integrate with respect to \(y\) first. Lets take a quick look at an example of this. So, to do this well need a parameterization of \(C\). and because it is also times \(g\left( x \right)\) we can factor the \(h\left( y \right)\) out of the inner integral. However, if we cut the disk in half and rename all the various portions of the curves we get the following sketch. Next, use Greens theorem on each of these and again use the fact that we can break up line integrals into separate line integrals for each portion of the boundary. In this case well integrate with respect to \(x\) first. Finally using the second portion of the definition of the inverse tangent function gives us. and see if we can get some functions \(P\) and \(Q\) that will satisfy this. As we traverse each boundary the corresponding region is always on the left. This is the only indefinite integral in this section and by now we should be getting pretty good with these so we wont spend a lot of time on this part. Welcome to my math notes site. Doing this gives. To do this we need to recall the definition of absolute value. Note that you are NOT asked to find the solution only show that at least one must exist in the indicated interval. They were created by Khan Academy math experts and reviewed for curriculum alignment by experts at both Illustrative Mathematics and Khan Academy. We will discuss factoring out the greatest common factor, factoring by grouping, factoring quadratics and factoring polynomials with degree greater than 2. Get 247 customer support help when you place a homework help service order with us. In fact we can say more. First, determine where the quantity inside the absolute value bars is negative and where it is positive. This is the last topic that we need to discuss in this section. Again, if youd like to verify this a quick sketch of a unit circle should convince you that this range will cover all possible values of cosine exactly once. This is shown below. Step - 4: Also, represent all excluded values on the number line using open circles. As with the inverse sine we are really just asking the following. We will discuss several methods for determining the absolute minimum or maximum of the function. Finally, also note that we can think of the whole boundary, \(C\), as. Lets first sketch \(C\) and \(D\) for this case to make sure that the conditions of Greens Theorem are met for \(C\) and will need the sketch of \(D\) to evaluate the double integral. Note that this section is only intended to introduce these concepts and not teach you everything about them. Another way to think of a positive orientation (that will cover much more general curves as well see later) is that as we traverse the path following the positive orientation the region \(D\) must always be on the left. It is important to understand the difference between the two types of minimum/maximum (collectively called extrema) values for many of the applications in this chapter and so we use a variety of examples to help with this. With the Mean Value Theorem we will prove a couple of very nice facts, one of which will be very useful in the next chapter. Now, in the first integral we have \(t < \frac{5}{3}\) and so \(3t - 5 < 0\) in this interval of integration. In the previous examples where we had functions that werent continuous we had division by zero and no matter how hard we try we cant get rid of that problem. As with the inverse sine weve got a restriction on the angles, \(y\), that we get out of the inverse cosine function. So, not too bad of an integral there provided you get the substitution. Notice that this is the same line integral as we looked at in the second example and only the curve has changed. We think of all the \(x\)s as constants and integrate with respect to \(y\) or we think of all \(y\)s as constants and integrate with respect to \(x\). Next, notice that because the inner integral is with respect to \(x\) and \(h\left( y \right)\) is a function only of \(y\) it can be considered a constant as far as the \(x\) integration is concerned (changing \(x\) will not affect the value of \(y\)!) In the first integral we will have \(x\) between -2 and 1 and this means that we can use the second equation for \(f\left( x \right)\) and likewise for the second integral \(x\) will be between 1 and 3 and so we can use the first function for \(f\left( x \right)\). We can identify \(P\) and \(Q\) from the line integral. Sure enough the same answer as the first solution. and divide every term by cos2 \(y\) we will get. To this point weve not seen any functions that will differentiate to get an absolute value nor will we ever see a function that will differentiate to get an absolute value. The integrand in this case is odd and the interval is in the correct form and so we dont even need to integrate. There are many functions that will satisfy this. For this integral well integrate with respect to \(y\) first. So, weve computed a fair number of definite integrals at this point. So, using Greens Theorem the line integral becomes. However, there are many functions out there that arent zero when evaluated at zero so be careful. The graph reveals a problem. This should explain the similarity in the notations for the indefinite and definite integrals. Rational Inequalities; Absolute Value Equations; Absolute Value Inequalities; Graphing and Functions. So, Greens theorem, as stated, will not work on regions that have holes in them. Step - 3: Represent all the values on the number line. In mathematics, the absolute value or modulus of a real number, denoted | |, is the non-negative value of without regard to its sign.Namely, | | = if x is a positive number, and | | = if is negative (in which case negating makes positive), and | | =. Without them we couldnt have done the evaluation. Here is a set of practice problems to accompany the Absolute Value Inequalities section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University. Note that there are in fact two ways of computing a double integral over a rectangle and also notice that the inner differential matches up with the limits on the inner integral and similarly for the outer differential and limits. If \(f\left( x \right)\) is an even function then. Dont forget your basic Calculus I substitutions! The Shape of a Graph, Part II In this section we will discuss what the second derivative of a function can tell us about the graph of a function. In this section however, we will need to keep this condition in mind as we do our evaluations. Please contact Savvas Learning Company for product support. This will give a function involving only \(x\)s which we can in turn integrate. So, using the fact cut the evaluation in half (in essence since one of the new limits was zero). Recall from our first example above that all we really need here is any anti-derivative of the integrand. The tangent and inverse tangent functions are inverse functions so, Therefore, to find the derivative of the inverse tangent function we can start with. Note that the absolute value bars on the logarithm are required here. In the following sets of examples we wont make too much of an issue with continuity problems, or lack of continuity problems, unless it affects the evaluation of the integral. Dont forget to convert the radical to fractional exponents before using the product rule. In this case the region \(D\) will now be the region between these two circles and that will only change the limits in the double integral so well not put in some of the details here. However, sometimes one direction of integration is significantly easier than the other so make sure that you think about which one you should do first before actually doing the integral. Prop 30 is supported by a coalition including CalFire Firefighters, the American Lung Association, environmental organizations, electrical workers and businesses that want to improve Californias air quality by fighting and preventing wildfires and reducing air pollution from vehicles. We will revisit finding the maximum and/or minimum function value and we will define the marginal cost function, the average cost, the revenue function, the marginal revenue function and the marginal profit function. Printable in convenient PDF format. Now, on some level this is just notation and doesnt really tell us how to compute the double integral. The examples in this section tend to be a little more involved and will often involve situations that will be more easily described with a sketch as opposed to the 'simple' geometric objects we looked at in the previous section. Take the last integral as an example. An odd function is any function which satisfies. Step - 2: Solve the equation for one or more values. Its generally easier to evaluate the term with positive exponents. In particular we got rid of the negative exponent on the second term. What we want to do is discuss single indefinite integrals of a function of two variables. Now lets take a look at the inverse cosine. Based on this graph determine where the function is discontinuous. With the chain rule in hand we will be able to differentiate a much wider variety of functions. In this case because the limits for \(x\) are kind of nice (i.e. For the most part answering these questions isnt that difficult. In this section we are going to look at the derivatives of the inverse trig functions. Now, \(\int_{a}^{b}{{g\left( x \right)\,dx}}\) is a standard Calculus I definite integral and we know that its value is just a constant. From a unit circle we can see that we must have \(y = \frac{{3\pi }}{4}\). The integral is. First, notice that we will have a division by zero issue at \(w = 0\), but since this isnt in the interval of integration we wont have to worry about it. We will work a number of examples illustrating how to find them for a wide variety of functions. This notation is, You appear to be on a device with a "narrow" screen width (, \[\begin{array}{ll}\displaystyle \frac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \frac{1}{{\sqrt {1 - {x^2}} }} & \hspace{1.0in}\displaystyle \frac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right) = - \frac{1}{{\sqrt {1 - {x^2}} }}\\ \displaystyle \frac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \frac{1}{{1 + {x^2}}} & \hspace{1.0in}\displaystyle \frac{d}{{dx}}\left( {{{\cot }^{ - 1}}x} \right) = - \frac{1}{{1 + {x^2}}}\\ \displaystyle \frac{d}{{dx}}\left( {{{\sec }^{ - 1}}x} \right) = \frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }} & \hspace{1.0in}\displaystyle \frac{d}{{dx}}\left( {{{\csc }^{ - 1}}x} \right) = - \frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }}\end{array}\], Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, \(f\left( t \right) = 4{\cos ^{ - 1}}\left( t \right) - 10{\tan ^{ - 1}}\left( t \right)\), \(y = \sqrt z \, {\sin ^{ - 1}}\left( z \right)\). 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In turn integrate now, on some level this is here only make. ( x\ ) first out this section 2: Solve the equation for one or more values, by... Minimum or maximum of the inverse trig functions to convert the radical fractional. You get the following we talk about double absolute value inequalities anti-derivative for a wide variety functions! \Right ) \ ) same answer as the result of using Greens,! That have holes in them this is just notation and doesnt really us!
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