Find the most general antiderivative of:
3squarerootx - 8/x^9
I get confused about the squareroot..
2007-12-02 10:42:27 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I = 3 ∫ x^(1/2) dx - 8 ∫ x^(- 9) dx
I = (2) x^(3/2) - 8 x^(- 8) /(- 8 ) + C
I = (2) x^(3/2) + 1 / x^(8) + C
2007-12-04 10:00:13 · answer #1 · answered by Como 7 · 2⤊ 0⤋
first, I changed the rad(x) to x^1/2 so then I could easily see that the exponent should be 3/2. And I also changed the ^9 to ^-9 so then it was easy to see that the exponent should be -8. If I made a mistake, i've been out of high school for 2.5 years.
3x^1/2 - 8x^-9
2x^3/2 + 1/x^8 + c
2007-12-02 18:47:53 · answer #2 · answered by David 4 · 0⤊ 0⤋
Think of a square root as something to the (1/2) power.
Answer is 2x^(3/2) + 1/x^8.
2007-12-02 18:47:10 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Just split it into 2 parts(assuming x is the only thing in the square root.
Then for 2*root(x) raise the power by 1 so you have 3x^1.5 and divide by the power(1.5) to get 2x^1.5. For the other part do the same to get -(-1)/x^8 so the whole thing is 2x^1.5 +/x^8
2007-12-02 18:48:38 · answer #4 · answered by smithersprime 4 · 0⤊ 0⤋
1/x(1/x^8)= 1/x^9
2007-12-02 18:46:02 · answer #5 · answered by Anonymous · 0⤊ 0⤋