Tutor profile: Stephanie R.
Obtain the integral of (x)(e^x)dx.
We have: Integral of (x)(e^x)dx. Note that the variable x is changing in two functions ((x) is an algebraic function and (e^x) is an exponential function). Therefore we recognize that we must use integration by parts to solve for the integral. Although this may be tedious, there are easy tricks to memorize that will make the math simple. Remember this equation like a song: "integral of udv is equal uv minus integral of vdu." Note here that on the left side of the equation, "u" is assigned to one function and "dv" is assigned to another function in our problem. How do we choose? Memorize this abbreviation: ILATE. I=inverse trigonometric function. L=logarithmic function. A=algebraic function. T=trigonometric function. E=exponential function. This abbreviation will help you choose what "u" will be. (x) in our problem is listed as A. (e^x) is listed as E. A comes before E in our abbreviation, so we choose "u" to be (x). Thus, "dv" will be assigned to (e^x). Now, determine "du" and "v" to complete our equation. "du" is dx (since we are getting the derivative of "u", which we assigned to be x). "v" is e^x (since we are getting the antiderivative of "dv", which we assigned to be e^x). With all values known, place them into the right side of the expression, which is: uv-integral of vdu. We get: (x)(e^x)-integral of [(e^x)(dx)]. Complete the integral term to get: (x)(e^x)- (e^x) +C. Since we do not have finite bounds, DON'T FORGET YOUR PLUS C (some constant)! Therefore, using integration by parts states that the Integral of (x)(e^x)dx is equal to: (x)(e^x)- (e^x) +C.
A farmer needs to fence in his land. One side is already fenced in. He has 300 ft of fencing to use. What dimensions (x and y) will optimize the area enclosed by the fence?
This is an optimization problem. First, understand the situation. We want x and y values (two changing variables) that will result in the largest possible area. We also have one degree of freedom in the problem: the amount of fencing is fixed at 300 ft. We must work with this constraint. Next, formulate the governing equations. We need an equation that explains the area (denoted as A) and another for the perimeter enclosing the area. The equations are as follows: A=xy ; 300ft=x+2y. (Note that since one side is already fenced in, we have only one x. If this were not the case, the equation would become 300=2x+2y.) We have 2 unknowns and 2 governing equations: the problem is solvable. We have two changing variables, so let's solve for x in terms of y. We get: x=300-2y. Now, we can substitute this expression into the equation for the area. We get: A=(300-2y)(y), or A=300y-2y^2. Now we have successfully defined the equation in terms of one variable while incorporating our constraint for the perimeter of the fence. Optimizing the area means obtaining the derivative of the area and setting the area equal to zero. By doing this, we are finding the critical point (the max point of the expression) that will give us the max possible value of y. We get: 0=300-4y. Solving for y, we get y=75. Earlier, we defined x in terms of y to be x=300-2y. Solving for x gives us: x=150. We found the dimensions, now let's find the area. A=(150)(75) ==> A=11250ft^2. This should be the max area. However, always CHECK your calculations. Does x and y sum to the perimeter of 300? 300=(150)+2(75) ==> yes! Also check if these dimensions make sense. We can find our bounds of y (or intervals) by setting x=0. We get y=150. Set y=0 and we get x=300. The bounds of y are [0,150]. This tells us that the max value of y lies between this interval, which it does.
How would you obtain the radius of 1 mole of ideal gas particles exposed to a pressure of 1 atm and a temperature of 25 degrees Celsius? Note that the value of the ideal gas constant is equal to 0.0821 (L)(atm)/[(mol)(K)]
First, define the problem statement. Understand what the problem is asking you to find, and write down a list of all known information and unknown information . Unknown: r=radius of the ideal gas particle. Known: we know that the gas particle is ideal, so we can use the ideal gas equation (this equation is purely chemistry based), which is PV=nRT (Pressure)(Volume)= (number of moles of gas)(Ideal gas constant)(Temperature). Note that we must assume that the gas particle is spherical. We know that the volume of a sphere is (4/3)pi(r^3). r is the unknown variable. Substitute all the values into the ideal gas equation. Isolate r to solve for it in terms of given variables. BEFORE YOU SOLVE, you must not forget to convert all of the units! Convert liters to meters to get the r in units of meters (1000L=1m^3). Also, in scientific equations, we work in absolute temperatures, so convert Celsius to Kelvin (25 degrees Celsius+273). You must get the cube root of the equation to obtain r.
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