First, you should solve for
, which equals
. Now, solve the integral of
=
, to get that
. You can check this by taking the integral of what you got. Now by the Fundamental Theorem
![\int\limits^2_0 {4x} \, dx=[2x^2] ^{2}_{0}=2(2)^{2}-2(0)^2=8](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E2_0%20%7B4x%7D%20%5C%2C%20dx%3D%5B2x%5E2%5D%20%5E%7B2%7D_%7B0%7D%3D2%282%29%5E%7B2%7D-2%280%29%5E2%3D8)
.
This should be the answer to your question, if I understood what you were asking correctly.
Answer:
the radius of circle s is point,radius,arc,
Answer: {(x + 2), (x - 1), (x - 3)}
Step-by-step explanation:
Presented symbolically, we have:
x^3 - 2x^2 - 5x + 6
Synthetic division is very useful for determining roots of polynomials. Once we have roots, we can easily write the corresponding factors.
Write out possible factors of 6: {±1, ±2, ±3, ±6}
Let's determine whether or not -2 is a root. Set up synthetic division as follows:
-2 / 1 -2 -5 6
-2 8 -6
-----------------------
1 -4 3 0
since the remainder is zero, we know for sure that -2 is a root and (x + 2) is a factor of the given polynomial. The coefficients of the product of the remaining two factors are {1, -4, 3}. This trinomial factors easily into {(x -1), (x - 3)}.
Thus, the three factors of the given polynomial are {(x + 2), (x - 1), (x - 3)}
