The integration of functions is the opposite derivatives so they can be used to find the antiderivative of a function. This is use to find the area above/below x aixs of the curve.
(CueMath)
| Formula | Explain | Example 1 | Example 2 |
|---|---|---|---|
| $$\int x^ndx = \frac{x^{n+1}}{n+1} +C$$ | Add one to the power then divide by the old power plus 1 and lastly plus C at the end | $$\int x^2 dx==> \frac {x^3}{3} +C$$ | $$\int66x^{10} dx==> 6x^{11} +C$$ |
| $$\int dx = x + C$$ | When it just dx, you add x to that number | $$\int 2 dx==> 2x + c$$ | $$\int 10 dx==> 10x + c$$ |
| $$\int \cos x\: dx = \sin x + C$$ | cos turns into sin | $$\int \cos 0 \: dx = \sin0 + C$$ | $$\int \cos 90 \: dx = \sin90 + C$$ |
| $$\int \sin x \: dx = -\cos x + C$$ | sin turns into -cos | $$\int \sin 0 \: dx = -\cos(0) + C$$ | $$\int \sin 90 \: dx = -\cos(90) + C$$ |
| $$\int (trig^1)\: k x \: dx = \frac{(trig^1) \: kx}{k} + C$$ | (trig^1) is the any trigonometric being use above as this formula can be used more than once. K is the extra number next to x that you will have to divide by all other then c |
$$\frac{d}{dx} b^5 = b^5\ln(b) $$ | $$\frac{d}{dx} b^45 = b^4\ln(b) $$ |
| $$\frac{d}{dx} \ln(x) = \frac{1}{x}$$ | Log turns into fraction x of 1 | $$\frac{d}{dx} \ln(5) = \frac{1}{5}$$ | $$\frac{d}{dx} \ln(45) = \frac{1}{45}$$ |
Example 1, definite integrals of $$\int^3_0 7x \: dx $$
Using the first rule in the table from above.: $$\int x^ndx = \frac{x^{n+1}}{n+1} +C$$
$$\int^3_0 7x \: dx ===> [\frac {7x^2}{2}]^3_0 +C $$
Now, substitute the function, do the range and ignore the c for now (the 3 and the 0)
$$[\frac {7x^2}{2}]^3_0 +C ===> \frac {7(3)^2}{2} - \frac {7(0)^2}{2} $$
Then solve = 31.5 - 0 = 31.5
Therefore x = 31.5
Click for exact equation = https://www.desmos.com/calculator/f7gcw5cmr6
The the green curve show the equation as a qudratic and the red is the 3 and 0 x axis that show the underneath area
There are quiz at https://www.bbc.co.uk/bitesize/guides/zgxttfr/test
Click for referance(bitesize)