Could somebody please explain how i can integrate these, i have tried but i just get lost.
1)$\displaystyle \int(2+0.5x-6x^2)dx$
2) $\displaystyle \int(x-x^{-2}-x^{0.5})dx$
Could somebody please explain how i can integrate these, i have tried but i just get lost.
1)$\displaystyle \int(2+0.5x-6x^2)dx$
2) $\displaystyle \int(x-x^{-2}-x^{0.5})dx$
$\displaystyle \int{2+\frac{x}{2}-6x^2}dx$...so using the general rule you get $\displaystyle 2x+frac{x^2}{4}-2x^3+C$
and using the same method Mr. fantastic showed you you get the antiderivative of the other to be $\displaystyle \frac{x^2}{2}+\frac{1}{x}-\frac{2x^{\frac{3}{2}}}{3}+C$
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:
$\displaystyle \int af(x) \, dx = a\int f(x) \, dx $
And, you can break it up and integrate the smaller parts, for example:
$\displaystyle \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:
$\displaystyle
\int x^n \, dx = \frac{1}{n+1} x^{n+1}, \, n \neq -1
$
for example:
Integrate:
$\displaystyle \int 2x^2 + 4x + 3 \, dx$
$\displaystyle \int 2x^2 \, dx + \int 4x \, dx + \int 3 \, dx$
I broke it up into the integration of separate terms.
$\displaystyle 2\int x^2 \, dx + 4\int x \, dx + \int 3 \, dx$
Then I isolated the constant in each term.
Now we begin integration
$\displaystyle 2 (\frac {1}{2+1} x^{2+1}) + 4 (\frac {1}{1+1} x^{1+1}) + 3x + C$
Here, we used the antiderivative power rule, as described by Mr. F and this simplifies to:
$\displaystyle \frac {2}{3}x^3 + 2x^2 + 3x + C$
And that's the integral.
Also, think in terms of fractions $\displaystyle 0.5 = \frac{1}{2}$ for the second problem, it will help enormously. And don't forget the plus C! very important.
Hope this helps!