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Evaluate the line integral where C is the given curve. We're integrating over the curve C, y to the third ds, and C is the curve with parametric equations x = t cubed, y = t. We're going from t = 0 to t = 2. So we're going to integrate over that curve C of y to the third ds. We're going to convert everything into our parameter t in terms of our parameter t. So I'm going to be integrating from t = 0 to t = 2. Those will be my limits of integration. Now y is equal to t, so I'm going to replace y with what it's equal to in terms of t. So I'm going to be integrating the function t to the third. Now ds we're going to write as a square root of dx dt squared + dy dt squared, squared of all that as we said dt. So we're integrating now everything with respect to t. So this is going to be equal to the integral from 0 to 2 of t to the third times the square root of -- see the derivative of x with respect to t is 3 t squared. So we have 3 t squared squared + dy dt; well, that's just 1 squared dt. So we have the integral from 0 to 2 of t to the third times the square root of 9 t to the fourth + 1 dt. So this is a pretty straightforward integration here. We're going to let u be equal to 9 t to the fourth + 1 then du is equal to 36 t to the third dt and so that tells me I can replace a t to the third dt with a du over 36. And so we're going to have the integral then from -- well, new limits of integration. I'm just going to put some squiggly marks there to remind myself that we switched variables. So I'm not going from t = 0 to t = 2. I'm doing things in terms of you right now. But I have a 1 over 36. I'll put that out front, and we're going to have the square root of u. So u to the 1/2, t to the third dt was replaced by du over 36. We got the 36 out front. And so now this is a pretty easy antiderivative in terms of u. It's u to the 3/2 times 2/3. And again, different limits of integration. We could figure out what they are in terms of u, but I'm going to convert back into t. So we're going to have 1 over 36 times 2/3 times u to the 3/2. Now, u is 9 t to the fourth + 1, that to the 3/2 power. And now we can go ahead and go from original limits of integration 0 to 2. So let's see, when I put a 2 in here, we're going to have -- 1 over 36 times 2/3. That's going to be 1 over 54, isn't it? So we'll have 1 over 54 times -- putting a 2 in, we have 9 times 2 to the fourth. That's 9 times 16, which is 144 + 1, is 145. So we put the 2 in there, we get 145 to the 3/2 minus, putting the 0 in, we get 9 x 0 to the fourth. That's 0. 0 + 1 is 1. So we just get 1 to the 3/2 or 1. So let's see, what's the best way to write this. How about 1 over 54 -- I guess we could leave it like that. We could also write 145 to the 3/2 as 145 times the square root of 145 and then minus 1. And that is that line integral of y to the third ds over the given curve C.

Evaluate the line integral where C is the given curve. We're integrating over the curve C, y to the third ds, and C is the curve with parametric equations x = t cubed, y = t. We're going from t = 0 to t = 2. So we're going to integrate over that curve C of y to the third ds. We're going to convert everything into our parameter t in terms of our parameter t. So I'm going to be integrating from t = 0 to t = 2. Those will be my limits of integration. Now y is equal to t, so I'm going to replace y with what it's equal to in terms of t. So I'm going to be integrating the function t to the third. Now ds we're going to write as a square root of dx dt squared + dy dt squared, squared of all that as we said dt. So we're integrating now everything with respect to t. So this is going to be equal to the integral from 0 to 2 of t to the third times the square root of -- see the derivative of x with respect to t is 3 t squared. So we have 3 t squared squared + dy dt; well, that's just 1 squared dt. So we have the integral from 0 to 2 of t to the third times the square root of 9 t to the fourth + 1 dt. So this is a pretty straightforward integration here. We're going to let u be equal to 9 t to the fourth + 1 then du is equal to 36 t to the third dt and so that tells me I can replace a t to the third dt with a du over 36. And so we're going to have the integral then from -- well, new limits of integration. I'm just going to put some squiggly marks there to remind myself that we switched variables. So I'm not going from t = 0 to t = 2. I'm doing things in terms of you right now. But I have a 1 over 36. I'll put that out front, and we're going to have the square root of u. So u to the 1/2, t to the third dt was replaced by du over 36. We got the 36 out front. And so now this is a pretty easy antiderivative in terms of u. It's u to the 3/2 times 2/3. And again, different limits of integration. We could figure out what they are in terms of u, but I'm going to convert back into t. So we're going to have 1 over 36 times 2/3 times u to the 3/2. Now, u is 9 t to the fourth + 1, that to the 3/2 power. And now we can go ahead and go from original limits of integration 0 to 2. So let's see, when I put a 2 in here, we're going to have -- 1 over 36 times 2/3. That's going to be 1 over 54, isn't it? So we'll have 1 over 54 times -- putting a 2 in, we have 9 times 2 to the fourth. That's 9 times 16, which is 144 + 1, is 145. So we put the 2 in there, we get 145 to the 3/2 minus, putting the 0 in, we get 9 x 0 to the fourth. That's 0. 0 + 1 is 1. So we just get 1 to the 3/2 or 1. So let's see, what's the best way to write this. How about 1 over 54 -- I guess we could leave it like that. We could also write 145 to the 3/2 as 145 times the square root of 145 and then minus 1. And that is that line integral of y to the third ds over the given curve C.

A linear equation is an equation in which each term is either a constant or the product of a constant times the first power of a variable. In simple terms it is a mathematical sentence in which you can see only one letter (which might appear more than once) but there will be no powers (squared, cubed etc). Here is an example of a simple linear equation:

2x + 7 = 15

This equation can be "solved" to find which value is represented by the letter x.

The eQuation Generator above can make up unlimited equations for you to practise solving. You can change the options so that one of five different types of equation is displayed. It is not possible to predict how quickly you will develop confidence solving equations of a particular type but typically the examples will increase in difficulty very slightly each time you press the Next button. A Restart button is provided if the questions generated start to become a little too difficult. This button restarts the difficulty level but will present different equations.

Here are examples showing a good way to solve equations by thinking of the two sides of the equation as two sides of a balance. The equation will remain balanced only if you do the same thing (multiply, divide add or subtract) to both sides.

2(3x−4)+1=52(3x−4)+1=5
Subtract 1 from both sides
2(3x−4)=42(3x−4)=4
Divide both sides by 2
3x−4=23x−4=2
Add 4 to both sides
3x=63x=6
Divide both sides by 3
x=2x=2

Another method:

2(3x−4)+1=52(3x−4)+1=5
First multiply out (expand) the brackets
6x−8+1=56x−8+1=5
Collect together like terms
6x−7=56x−7=5
Add 7 to both sides
6x=126x=12
Divide both sides by 6
x=2

2x+35+7=3x+1232x+35+7=3x+123
Multiply both sides by 15 (the lowest common multiple of the denominators)
6x+9+105=15x+606x+9+105=15x+60
Collect together like terms
6x+114=15x+606x+114=15x+60
Subtract 60 from both sides
6x+54=15x6x+54=15x
Subtract 6x from both sides
54=9x54=9x
Divide both sides by 9
x=6

1. 12+14+18+116+132+⋯=112+14+18+116+132+⋯=1. This is a geometric sequence that converges to 1.

2. eπi=−1eπi=−1 Very famous identity discovered by Euler.

3. sin2(θ)+cos2(θ)=1sin2(θ)+cos2(θ)=1 A basic trig identity.

4. V−E+F=1V−E+F=1 where V is the number of vertexes, E is the number of edges, F is the number of faces of a polyhedron.

5. (∑∞m=1μ(m)ms)(∑∞n=11ns)=1(∑m=1∞μ(m)ms)(∑n=1∞1ns)=1. Where μ(m)μ(m) is the Mobius function. This is not as famous, it states that the Dirichlet series with numerator μ(m)μ(m) inverts the one with numerator 1(which is the Riemann Zeta function).