The formula for angular velocity: v = w r
v / w = r ( true )
v / wr = 1 ( true )
v / r = w ( true )
Not equivalent is : A ) v W = r
Answer:
-2(n-3)
Step-by-step explanation:
Hi there! Assume that this is your question.
![\large{ \int \limits^a_b ( {x}^{2} + 2x)dx}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Cint%20%5Climits%5Ea_b%20%28%20%7Bx%7D%5E%7B2%7D%20%20%2B%202x%29dx%7D)
Before we get to Integral, you have to know Differentiation first. If you know how to differentiate a polynomial function then we are good to go in Integral!
We call the function that we are going to integrate as Integrand. Integrand is a function that's differentiated. In Integral, Integrating requires you to turn the function from differentiated to an original function.
For Ex. If the Integrand is x² then the original function is (1/3)x³ because when we differentiate (1/3)x³, we get x²
![\large{f(x) = \frac{1}{3} {x}^{3} \longrightarrow f'(x) = {x}^{2} } \\ \large{f'(x) = 3( \frac{1}{3} ) {x}^{3 - 1} } \\ \large{f'(x) = {x}^{2} }](https://tex.z-dn.net/?f=%20%5Clarge%7Bf%28x%29%20%3D%20%20%5Cfrac%7B1%7D%7B3%7D%20%20%7Bx%7D%5E%7B3%7D%20%20%5Clongrightarrow%20f%27%28x%29%20%3D%20%20%7Bx%7D%5E%7B2%7D%20%7D%20%5C%5C%20%20%20%5Clarge%7Bf%27%28x%29%20%3D%203%28%20%5Cfrac%7B1%7D%7B3%7D%20%29%20%7Bx%7D%5E%7B3%20-%201%7D%20%7D%20%5C%5C%20%20%5Clarge%7Bf%27%28x%29%20%3D%20%20%7Bx%7D%5E%7B2%7D%20%7D)
So when we Integrate, make sure to convert Integrand as in original function. From the question, our Integrand is x²+2x. The function is in differentiated form. We know that x² is from (1/3)x³ and 2x comes from x²
![\large{ f(x) = {x}^{2} \longrightarrow f'(x) = 2x} \\ \large{f'(x) = 2 {x}^{2 - 1} } \\ \large{f'(x) = 2x}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20f%28x%29%20%3D%20%20%7Bx%7D%5E%7B2%7D%20%20%5Clongrightarrow%20f%27%28x%29%20%3D%202x%7D%20%5C%5C%20%20%5Clarge%7Bf%27%28x%29%20%3D%202%20%7Bx%7D%5E%7B2%20-%201%7D%20%7D%20%20%5C%5C%20%20%5Clarge%7Bf%27%28x%29%20%3D%202x%7D)
Thus,
![\large{ \int \limits^a_b ( {x}^{2} + 2x)dx} \\ \large{\int \limits^a_b ( \frac{1}{3} {x}^{3} + {x}^{2}) }](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Cint%20%5Climits%5Ea_b%20%28%20%7Bx%7D%5E%7B2%7D%20%20%2B%202x%29dx%7D%20%20%20%5C%5C%20%20%5Clarge%7B%5Cint%20%5Climits%5Ea_b%20%28%20%5Cfrac%7B1%7D%7B3%7D%20%20%7Bx%7D%5E%7B3%7D%20%20%2B%20%20%7Bx%7D%5E%7B2%7D%29%20%7D)
Normally, if it's an indefinite Integral then we'd just put + C after (1/3)x³+x² but since we have a and b, it's a definite Integral.
![\large{ \int \limits^b_a f(x)dx = F(b) - F(a)}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Cint%20%5Climits%5Eb_a%20f%28x%29dx%20%3D%20F%28b%29%20-%20F%28a%29%7D)
Define F(x) as our anti-diff
From our problem, substitute x = a in then subtract with the one that substitute x = b
![\large{ (\frac{1}{3}{a}^{3} + {a}^{2} ) - ( \frac{1}{3} {b}^{3} + {b}^{2}) }](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%28%5Cfrac%7B1%7D%7B3%7D%7Ba%7D%5E%7B3%7D%20%2B%20%20%7Ba%7D%5E%7B2%7D%20%29%20-%20%28%20%5Cfrac%7B1%7D%7B3%7D%20%7Bb%7D%5E%7B3%7D%20%20%2B%20%20%7Bb%7D%5E%7B2%7D%29%20%20%7D)
Simplify as we get:
![\large \boxed{ \frac{1}{3}{a}^{3} + {a}^{2} - \frac{1}{3} {b}^{3} - {b}^{2}}](https://tex.z-dn.net/?f=%20%5Clarge%20%5Cboxed%7B%20%5Cfrac%7B1%7D%7B3%7D%7Ba%7D%5E%7B3%7D%20%2B%20%20%7Ba%7D%5E%7B2%7D%20%20-%20%20%5Cfrac%7B1%7D%7B3%7D%20%7Bb%7D%5E%7B3%7D%20%20%20-%20%20%20%7Bb%7D%5E%7B2%7D%7D)
Answer:
f(6) = -114
Step-by-step explanation:
f(x) = -4x² + 5x
f(6) = -4(6)² + 5(6)
f(6) = -4(36) + 30
f(6) = -114 + 30
f(6) = -114