All Calculus 1 Resources
Example Questions
Example Question #181 : Integral Expressions
Solve the following integral, where a and b are constants:
Keeping in mind that a and b are only constants, the integral is equal to
and was found using the following rule:
Example Question #182 : Integral Expressions
Evaluate the following integral:
To evaluate the integral, we must make the following substitution:
,
The derivative was found using the following rule:
Rewrite the integral in terms of u and integrate:
The integral was found using the following rule:
Finally, replace u with our original term:
.
Example Question #2262 : Calculus
In circuits with a resistor, the equation for voltage drop is given by:
, where is voltage, is charge, and is resistance.
Write the equation as an integral expression for .
Although this may seem really difficult, we only need to solve for .
To solve for , integrate both sides:
Example Question #183 : Integral Expressions
When integrating, remember to add one to the exponent and then put that result on the denominator: . Now evaluate at 2, and then 0. Then subtract the two results. .
Example Question #184 : Integral Expressions
The first step here is to chop this up into three separate terms and then simplify since we have only one denominator: . Then, integrate each term, remembering to add one to the exponent and then put that result on the denominator: . Simplify to get your answer: . Remember to add C because it is an indefinite integral.
Example Question #185 : Integral Expressions
First, chop this expression up into two terms: . Then, integrate each term, remembering that when there is a single x on a denominator, the integral is . Therefore, the integration is: . Remember to add C because it is an indefinite integral.
Example Question #186 : Integral Expressions
First, integrate each term separately. Remember, when integrating, raise the exponent by one and then also put that result on the denominator: . Then evaluate at 3 and then 0. Subtract two results to get: .
Example Question #187 : Integral Expressions
To integrate this expression, I would first rewrite it to be . Then, add one to the exponent and then put that result on the denominator: . Then, simplify to get your answer of . Remember to add C because it is an indefinite integral.
Example Question #1 : Other Writing Equations
Given the one-to-one equation f(x)=3x+1, the inverse function f-1(y)=
(y+1)/3
3x-1
(x-1)/3
(y-1)/3
undefined
(y-1)/3
Before we solve the problem by computation, let's look at the answer choices and see if we can eliminate any answer choices. We know that an inverse function must exist because f(x) is one-to-one, so we can eliminate the answer choice "undefined." Next, we know that the inverse function has to be in terms of y, so we can eliminate the two answer choices with an "x."
Now we can look at the two remaining answer choices. Let y=f(x) and solve for x to find the inverse.
So f(x)=y=3x+1. Solve for x.
x=(y-1)/3
Therefore our answer is (y-1)/3.
Example Question #1 : Other Writing Equations
Describe the graph of the polar equation r=-6.
The polar equation formula here is x2+y2=r2.
This is the formula for a cirlce, so we can eliminate the two "line" answers.
Plugging in r=-6 into the equation gives x2+y2=36, which describes the graph of a circle with radius 6.
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