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The derivative of the function
f is defined by
f^(')(x)=x^(2)-4x+2cos(2x-4). If
f(0)=1, then use a calculator to find the value of
f(6) to the nearest thousandth.
Answer:

The derivative of the function $f$ is defined by $f^{\prime}(x)=x^{2}-4 x+2 \cos (2 x-4)$ . If $f(0)=1$ , then use a calculator to find the value of $f(6)$ to the nearest thousandth. $\newline$ Answer:

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Evaluate variable expressions for sequences

Full solution

Q. The derivative of the function

f

is defined by

f^{\prime}(x)=x^{2}-4 x+2 \cos (2 x-4)

. If

f(0)=1

, then use a calculator to find the value of

f(6)

to the nearest thousandth.

\newline

Answer:

Set up integral: To find the value of $f(6)$ , we need to integrate the derivative $f'(x)$ from $0$ to $6$ and then add the initial value $f(0)$ to the result of the integration.
Integrate each term: First, we set up the integral of $f'(x)$ from $0$ to $6$ : $\int_{0}^{6} (x^2 - 4x + 2\cos(2x-4)) \, dx$
Evaluate integrals: We integrate each term separately: $\newline$ $\int_{0}^{6} x^2 \, dx = \left[\frac{x^3}{3}\right]_{0}^{6}$ $\newline$ $\int_{0}^{6} -4x \, dx = \left[-2x^2\right]_{0}^{6}$ $\newline$ $\int_{0}^{6} 2\cos(2x-4) \, dx = \left[\sin(2x-4)\right]_{0}^{6}$ (Note: We need to use substitution for this integral)
Substitute and integrate: We evaluate each integral at the bounds $0$ and $6$ :
$\left[\frac{x^3}{3}\right]$ (from $0$ to $6$ ) = $\left(\frac{6^3}{3}\right) - \left(\frac{0^3}{3}\right)$ = $72$
$\left[-2x^2\right]$ (from $0$ to $6$ ) = $6$ $0$ = $6$ $1$
For the cosine integral, we use substitution: let $6$ $2$ , then $6$ $3$ , so $6$ $4$ .
Evaluate cosine integral: Now we integrate $2\cos(u)$ with respect to $u$ : $\int 2\cos(u) \, du = 2\sin(u)$ We need to change the limits of integration according to the substitution: When $x = 0$ , $u = 2\cdot 0 - 4 = -4$ When $x = 6$ , $u = 2\cdot 6 - 4 = 8$ So we evaluate $2\sin(u)$ from $u = -4$ to $u = 8$ .
Add integrals: We find the value of the integral of the cosine term: $\newline$ $2\sin(u)$ (from $u = -4$ to $u = 8$ ) = $2\sin(8) - 2\sin(-4)$ $\newline$ Using a calculator, we find: $\newline$ $2\sin(8) \approx 2\times0.989 = 1.978$ $\newline$ $2\sin(-4) \approx 2\times(-0.756) = -1.512$ $\newline$ So the integral from $-4$ to $8$ is approximately $1.978 - (-1.512) = 3.490$
Add initial value: We add the results of the integrals: $72 - 72 + 3.490 = 3.490$
Add initial value: We add the results of the integrals: $72 - 72 + 3.490 = 3.490$ Finally, we add the initial value $f(0) = 1$ to the result of the integration to find $f(6)$ : $f(6) = 1 + 3.490 = 4.490$

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