Notation Note
S I’ll use this large “s” to indicate the elongated “s” which means “antiderivative”
[x] means “the absolute value of x”
Important Formulas
*1) S x^r dx = 1/(r+1) * x^(r+1) + C
*2) S e^kx dx = 1/k * e^kx + C
*3) S 1/x dx = ln [x] + C
****** Examples ********
#1
S 9x^8 dx
This type of problem uses formula *1) where r = 8
So applying the formula gives:
9 * 1/(8+1) * x ^ (8 + 1)
9 * (1/9) x^9
x^9 + C
#2
S e^3x dx
This type of problem uses formula *2) where k = 3.
So applying the formula:
S e^3x dx = 1/3 * e^3x + C
#3
S (5x – 7) ^2 dx
In this problem, we are going to treat the entire quantity: “(5x – 7)” as our “x” and use *1)
This gives us:
S (5x – 7)^2 dx = 1/(1+2)*(5x – 7)^(2+1)
= (1/3)(5x – 7) ^3 + C
However, we are not finished yet!
If you recall, if you take the derivative of the answer, you should get what you started with.
The derivative of (1/3)(5x – 7)^3 is (5x – 7)^2 * 5.
There is no “5” in our original problem, so we must get rid of it.
The way to “get rid of” the 5 is to multiply our answer by (1/5) because (1/5) * 5 = 1.
So our final answer is:
S (5x – 7)^2 dx = (1/3)(5x – 7)^3 * 5 + C
= (5/3)(5x – 7)^3 + C
#4
Find all functions f(x) with the following properties:
f'(x) = “square root of” x + 1 and f(4) = 0
First, we must find the antiderivative of f'(x).
In order to do that, the first thing we want to do is re-write f'(x) with powers, instead of the square root sign.
f'(x) = x ^(1/2) + 1
Now we find the antiderivative – using formula *1).
f(x) = 1/(1/2 +1) x^ (1/2 + 1) + x + C
= 1/(3/2) x ^(3/2) + x + C
= (2/3)* x^(3/2) + x + C
This is called the General Solution. But we want a more specific one, which means that we must find C.
To do that, we plug 4 in for x and set it equal to 0 (We know this because of the f(4) = 0) and then solve for C
0 = (2/3) * (4)^(3/2) + 4 + C
0 = (2/3)(8) + 4 + C
0 = 16/3 + 4 + C
0 = 16/3 + 12/3 + C
0 = 28/3 + C
C = -28/3
Now that we have found C, we replace the C in the anti-derivative that we already found.
f(x) = (2/3)* x^(3/2) + x – 28/3