Answer :
The solution of the definite integral from 1 to 3 of (3x² - 4x)dx is 10.
Work
Given the following integral: (3x² - 4x)dx and set boundaries of 1 and 3, we are asked to solve it.
Set up the definite integral.
[tex]\int\limits^3_1 {(3x^{2}-4x) } \, dx \\[/tex]
Integrate the function.
[tex]3x^2 --- > \frac{3x^3}{3} \\= x^3[/tex]
Applying the power rule where we add 1 to n and divide it by n + 1, we will get 3x³/3. Which simplifies to x³.
Doing the same thing for -4x, we will get:
[tex]-4x --- > \frac{-4x^2}{2} \\= -2x^2[/tex]
Now that we have our integral, we can substitute in the boundaries to solve for the areas. Make sure to subtract the lower boundary from the upper boundary.
[tex](x^3 - 2x^2) \\3^3 - 2(3^2) - 1^3 - 2(1^2) \\(27 - 18) - (1 - 2 ) \\9 - (-1) = 10[/tex]
Thus, the value of the integral with boundaries of x = 1, x = 3, is 10.
