1. ## Integrals

Could somebody please explain how i can integrate these, i have tried but i just get lost.

1) $\int(2+0.5x-6x^2)dx$

2) $\int(x-x^{-2}-x^{0.5})dx$

2. Originally Posted by ForgottenMemorie
$\int(2+0.5x-6x^2)dx$

$\int (x - x^{-2} - x^{0.5}) dx$ Mr F asks: Is this what you meant?
For both integrals, break them up into seperate integrals and use the rule $\int x^n \, dx = \frac{1}{n+1} x^{n+1}, \, n \neq -1$.

3. Originally Posted by mr fantastic
For both integrals, break them up into seperate integrals and use the rule $\int x^n \, dx = \frac{1}{n+1} x^{n+1}, \, n \neq -1$.
For example, for question 1: $\int x^2 \, dx = \frac{1}{2+1} x^{2+1},$

4. Originally Posted by ForgottenMemorie
For example, for question 1: $\int x^2 \, dx = \frac{1}{2+1} x^{2+1},$ Mr F adds: ${\color{red} = \frac{1}{3} x^3}$.
Hmmm well yes, I guess ....... in the sense that

$\int 2 + 0.5x - 6x^2 \, dx = \int 2 \, dx + 0.5 \int x \, dx - 6 \int x^2 \, dx$

and so you've done the last integral bit .....

5. I dont know how to do it, need help please.

6. Originally Posted by ForgottenMemorie
Could somebody please explain how i can integrate these, i have tried but i just get lost.

1) $\int(2+0.5x-6x^2)dx$

2) $\int(x-x^{-2}-x^{0.5})dx$
$\int{2+\frac{x}{2}-6x^2}dx$...so using the general rule you get $2x+frac{x^2}{4}-2x^3+C$

and using the same method Mr. fantastic showed you you get the antiderivative of the other to be $\frac{x^2}{2}+\frac{1}{x}-\frac{2x^{\frac{3}{2}}}{3}+C$

7. Hey,

I'm in Calc I right now, And I learned how to do this, so:

There's two rules to keep in mind when you work these:

Constant rule: You can kick a constant out of the integral to make it easier:

$\int af(x) \, dx = a\int f(x) \, dx$

And, you can break it up and integrate the smaller parts, for example:

$\int f(x) + g(x) + h(x) \, dx = \int f(x) \, dx + \int g(x) \, dx + \int h(x) \, dx$

And remember the antiderivative power rule that Mr. F presented:

$
\int x^n \, dx = \frac{1}{n+1} x^{n+1}, \, n \neq -1
$

for example:

Integrate:
$\int 2x^2 + 4x + 3 \, dx$

$\int 2x^2 \, dx + \int 4x \, dx + \int 3 \, dx$
I broke it up into the integration of separate terms.

$2\int x^2 \, dx + 4\int x \, dx + \int 3 \, dx$
Then I isolated the constant in each term.

Now we begin integration
$2 (\frac {1}{2+1} x^{2+1}) + 4 (\frac {1}{1+1} x^{1+1}) + 3x + C$
Here, we used the antiderivative power rule, as described by Mr. F and this simplifies to:

$\frac {2}{3}x^3 + 2x^2 + 3x + C$

And that's the integral.

Also, think in terms of fractions $0.5 = \frac{1}{2}$ for the second problem, it will help enormously. And don't forget the plus C! very important.

Hope this helps!