# Solve using the second funtamental theorem of calculus

1. Oct 17, 2009

### Jimmy84

1. The problem statement, all variables and given/known data
Solve using the second fundamental theorem of calculus
$$\int$$ from 0 to 2 of 2x^2 (√x^3 + 1) dx

2. Relevant equations

Using the second fundamental theorem of calculus

$$\int$$ from a to b of f(t) dt = g(b) - g(a)

3. The attempt at a solution

$$\int$$ from 0 to 2 of 2x^2 (√x^3 + 1)dx

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

This problem is solved in my book however I dont understand why the book added 2/3 and at the end of the equation (3x^2 dx) , in replace of 2x^2 .

Last edited: Oct 17, 2009
2. Oct 17, 2009

### VeeEight

Is this what you're asked for? $$\int$$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

3. Oct 18, 2009

### TheoMcCloskey

Jimmy84 -

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

4. Dec 12, 2009

### 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.

5. Dec 13, 2009

### HallsofIvy

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

Substitute u= x^3+ 1.

6. Dec 13, 2009

### Jimmy84

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

By the way it shouldnt 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 $$\int$$ from 0 to 2 of (√x^3 + 1) (3x^2 dx)