Solve using the second funtamental theorem of calculus

[Jimmy84]

Homework Statement

Solve using the second fundamental theorem of calculus
[tex]\int [/tex] from 0 to 2 of 2x^2 (√x^3 + 1) dx

Homework Equations

Using the second fundamental theorem of calculus

[tex]\int [/tex] from a to b of f(t) dt = g(b) - g(a)

The Attempt at a Solution

[tex]\int [/tex] from 0 to 2 of 2x^2 (√x^3 + 1)dx

= 2/3 multiplyed by [tex]\int [/tex] from 0 to 2 of (√x^3 + 1) (3x^2 dx)



This problem is solved in my book however I don't understand why the book added 2/3 and at the end of the equation (3x^2 dx) , in replace of 2x^2 .

 

[VeeEight]

Is this what you're asked for? [tex]\int [/tex]2x²(√ x³ + 1)dx

When I'm faced with a problem like this I usually take a guess at the function, differentiate it, and see how I can change it to get the proper answer. For example, say your guess is (√ x³ + 1). Differentiate it and see what you get - it's not the right answer but see what you can change

 

[TheoMcCloskey]

Jimmy84 -

This problem is solved in my book however I don't understand why the book added 2/3 and at the end of the equation (3x^2 dx) , in replace of 2x^2 .

Hint: Firstly, can you see that the two expressions for the integral are the same? Next, what is the derivative of [itex]x^3[/itex]. How can that be used in the new form of the integral expressed by your book.

 

[Jimmy84]

yea I can see that the two expressions for the integral are the same.

but I'm not sure at all where does the derivative of x^3 come from.

 

[HallsofIvy]

Actually, what's important is the derivative of x^3+ 1.

Substitute u= x^3+ 1.

 

[Jimmy84]

So it goes like this. u = x^3+ 1.

By the way it shouldn't be u = (the square root of x³ + 1) instead ?



Then dx is replaced by 3x^2 , and because of that 2/3 should be before the integral to keep the expresion the same in this way

2/3 multiplyed by [tex]\int [/tex] from 0 to 2 of (√x^3 + 1) (3x^2 dx)