You're going to have to back up, since your integrand is wrong. Show us what you did to get the part under the radical.jmiqbal said:Roughly: [tex]\int{\sqrt{1+144x}}\:dx = \frac{1}{216}(1 + 144x)^{3/2}[/tex]
miqbal said:I'm not the original poster dude.
[tex]s = \int_{a}^{b} \sqrt { 1 + [f'(x)]^2 }\, dx.[/tex]
[tex]f(x) = 5 + 8x^{\frac{3}{2}}[/tex]
[tex]f'(x) = 12x^{\frac{1}{2}}[/tex]
[tex][f'(x)]^{2} = 144x[/tex]
[tex]s = \int_{0}^{1} \sqrt { 1 + 144x }\, dx = {\frac{1}{216}(1 + 144x)^{3/2}}\left|^{1}_{0}[/tex]
To calculate y for 0 to 1, you will need to use the equation y = mx + b, where m is the slope and b is the y-intercept. You will also need to have a value for x, which represents the input value. Plug in the values for m, b, and x into the equation and solve for y. This will give you the corresponding y value for the input x.
Calculating y for 0 to 1 is commonly used in linear equations to find the value of the dependent variable (y) for a given independent variable (x). This can help in understanding the relationship between the two variables and making predictions based on the data.
No, this guide specifically focuses on calculating y for linear equations. For non-linear equations, the process will be different and may require other mathematical techniques.
If you don't have a value for the slope or y-intercept, you can use other methods such as graphing or finding the equation of a line to determine these values. Once you have the values, you can then follow the steps in this guide to calculate y for 0 to 1.
Yes, this guide can be used for any range of x as long as the equation is linear. However, if the values for x are very large or very small, it may be more accurate to use scientific notation or a computer program to perform the calculations.