Why would "negatives" not count? You can't ignore the abs value- you need to determine where x2 - 6x+ 3 is positive and where it is negative. Start by determining where x2- 6x+ 3= 0.weiseguy said:Homework Statement
[tex]\int^{8}_{0}\left|x^{2} - 6x + 3\right|dx[/tex]
This is for a single variable AP Calculus AB class in which we are solving using substitution method.
2. The attempt at a solution
I attempted it by just ignoring the abs value bars thinking that anything I am finding out "how much" of will be positive, and the negatives won't count. Solving with this in mind:
I came to the answer of 128/3
Please help as you can
You should have learned to solve quadratic equations long before a calculus class! I doubt that the part you missed was devoted to solving x2- 6x+ 3= 0! No, x= 3 is NOT a solution.weiseguy said:So (I haven't checked), if the interval 0 to 3 is negative and 3 to 8 is positive. I would say the answer is the negative interval plus the positive one?
I need a little further explanation I think. We just started this in class and I had to miss half of the lecture.
HallsofIvy said:You should have learned to solve quadratic equations long before a calculus class! I doubt that the part you missed was devoted to solving x2- 6x+ 3= 0! No, x= 3 is NOT a solution.
The integral of absolute value function is a mathematical calculation that determines the area under the curve of an absolute value function. It is represented by the symbol ∫|f(x)|dx and is equivalent to the sum of the positive and negative areas under the curve.
2.To find the integral of an absolute value function, you need to split the function into two separate parts: the positive part and the negative part. Then, you can integrate each part separately using the standard integration rules. Finally, you can combine the results to get the overall integral of the absolute value function.
3.The main difference between the integral of an absolute value function and a regular integral is that the former considers both the positive and negative areas under the curve, while the latter only considers the positive area. This is because the absolute value function always produces a positive result, even for negative inputs.
4.The integral of absolute value function has various real-life applications, such as calculating the distance traveled by an object with changing velocity, determining the total displacement of an object, and finding the average value of a function over a given interval. It is also used in economics, physics, and engineering to solve various problems involving distance, speed, and acceleration.
5.Unfortunately, there is no shortcut or easier method to find the integral of an absolute value function. It requires breaking the function into two parts and using standard integration rules. However, with practice and familiarity, the process can become more efficient and straightforward.