Use a taylor series to approximate the following definite integrals. retain as many terms as needed to ensure the error is less than 10−4. ∫0.201+x4√dx

I= 0.40102036

Step-by-step explanation:

∧ = raise to power

Given the equation

F(x) = √x² + 1.0

Now recalling the Taylor Series in respect to the approximation  for degree n at x = a

Therefore, it is given by :

f(x) = f(a) + f° (a) ((x - a)) + ( 1/2) f°° (a)((x -a)² + (1/n!) f∧n (a) ((x - a) ∧ n

Referring back to the Taylor Series

f(x) = - x ∧8/8 + x ∧ 4/2 + 1

If we integrate the Taylor series, we get,

f°(x) = - x ³*³/72 + x∧5/10 + x

At this point, we apply limit (lim) to the integrals

I = ∫ lim(0.4) (0) f(x) dx

I = ∫ (- x ∧8/8 + x ∧ 4/2 +1)dx

I = [ ( -x³*³/72 + x∧5/10 +x)]lim(0.4) lim(0)

The final answer then is:

I= 0.40102036

center (1,3) with radius 6

step-by-step explanation:

we must complete the square to find the center and radius of the circle.

first make sure the x and y squared terms have 1 as their coefficients. we also make sure x and y terms together.

we now create space between the x and y terms with parenthesis.

we complete the square by taking the middle terms -2x and the -6y - divide each and square them.

we add the squares to both sides.

simplify.

and write the quadratics in factored form.

the center is (h,k) or (1,3). the radius is the square root of 36 which is 6.

4

step-by-step explanation:

im pretty sure it's 4 beacsue 8/2 equals 4