English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories
0 ⤊

Given that dy/dx= (3x-2)^2 & that y=0 when x=1, calculate the value of y where x=1.5?

2007-09-24 03:10:25 · 1 answers · asked by hottie 1 in Science & Mathematics ➔ Mathematics

1 answers

We write this as dy = (3x - 2)^2dx and integrate to get y = [(3x - 2)^3]/9 + C. From y = 0 when x = 1 we find C = 1/9, so y = [(3x - 2)^3]/9 + 1/9. You do the evaluation at x = 1.5.

2007-09-24 04:09:02 · answer #1 · answered by Tony 7 · 0⤊ 0⤋

fedest.com, questions and answers