The derivative of the function f is defined by f^(')(x)=(x^(3)-5x)cos(2x). If f(-1)=-4, then use a calculator to find the value of f(6) to the nearest thousandth. Answer:

Integrate f^(')(x): To find f(6), we need to integrate f^(')(x) from -1 to 6 and add the result to f(-1).Apply integration by parts: The integral of f^(')(x) is ∫(x^(3)-5x)cos(2x) dx. We will use integration by parts and trigonometric identities to solve this.Integrate (1/2)sin(2x)(3x^(2)-5) dx: Let u = x^(3)-5x, then du = (3x^(2)-5) dx. Let dv = cos(2x) dx, then v = (1/2)sin(2x) using the integral of cosine.Apply integration by parts again: Using integration by parts, ∫udv = uv - ∫vdu. We apply this to our integral: ∫(x^(3)-5x)cos(2x) dx = (1/2)x^(3)sin(2x) - ∫(1/2)sin(2x)(3x^(2)-5) dx.Integrate -(3/2)x cos(2x) dx: We now need to integrate ∫(1/2)sin(2x)(3x^(2)-5) dx. This requires another integration by parts.Combine all parts of the integral: Let u = (3x^(2)-5), then du = 6x dx. Let dv = (1/2)sin(2x) dx, then v = -(1/4)cos(2x) using the integral of sine.Evaluate the definite integral: Applying integration by parts again, we get ∫(1/2)sin(2x)(3x^(2)-5) dx = -(1/4)(3x^(2)-5)cos(2x) - ∫-(1/4)cos(2x)(6x) dx.Calculate f(6): The remaining integral is ∫-(1/4)cos(2x)(6x) dx, which simplifies to ∫-(3/2)x cos(2x) dx. This requires another integration by parts.Let u = -3x, then du = -3 dx. Let dv = (1/2)cos(2x) dx, then v = (1/4)sin(2x).Applying integration by parts, we get ∫-(3/2)x cos(2x) dx = (1/4)(-3x)sin(2x) - ∫(1/4)(-3)sin(2x) dx.The remaining integral is ∫(1/4)(-3)sin(2x) dx, which simplifies to -(3/4) ∫sin(2x) dx. The integral of sin(2x) is -(1/2)cos(2x).So, ∫(1/4)(-3)sin(2x) dx = (3/8)cos(2x). Now we combine all parts of the integral we've calculated.The integral from -1 to 6 of f^(')(x) is the sum of all the parts we've integrated by parts. We need to evaluate this definite integral and add it to f(-1) = -4.We evaluate the integral at the bounds 6 and -1, subtract the lower bound evaluation from the upper bound evaluation, and add the result to f(-1) = -4.Using a calculator to evaluate the integral at x = 6 and x = -1, we find the change in f over this interval. We then add this change to f(-1) to find f(6).After calculating the definite integral and adding it to f(-1), we round the result to the nearest thousandth to get the final answer.

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The derivative of the function
f is defined by
f^(')(x)=(x^(3)-5x)cos(2x). If
f(-1)=-4, 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 $f^{\prime}(x)=\left(x^{3}-5 x\right) \cos (2 x)$ . If $f(-1)=-4$ , then use a calculator to find the value of $f(6)$ to the nearest thousandth. $\newline$ Answer:

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Algebra 2

Find trigonometric ratios using multiple identities

Full solution

Q. The derivative of the function

f

is defined by

f^{\prime}(x)=\left(x^{3}-5 x\right) \cos (2 x)

. If

f(-1)=-4

, then use a calculator to find the value of

f(6)

to the nearest thousandth.

\newline

Answer:

Integrate $f'(x)$ : To find $f(6)$ , we need to integrate $f'(x)$ from $-1$ to $6$ and add the result to $f(-1)$ .
Apply integration by parts: The integral of $f'(x)$ is $\int(x^{3}-5x)\cos(2x) \, dx$ . We will use integration by parts and trigonometric identities to solve this.
Integrate $(\frac{1}{2})\sin(2x)(3x^{2}-5) dx$ : Let $u = x^{3}-5x$ , then $du = (3x^{2}-5) dx$ . Let $dv = \cos(2x) dx$ , then $v = (\frac{1}{2})\sin(2x)$ using the integral of cosine.
Apply integration by parts again: Using integration by parts, $\int u\,dv = uv - \int v\,du$ . We apply this to our integral: $\int(x^{3}-5x)\cos(2x) \,dx = \frac{1}{2}x^{3}\sin(2x) - \int\frac{1}{2}\sin(2x)(3x^{2}-5) \,dx$ .
Integrate $-\frac{3}{2}x \cos(2x) \,dx$ : We now need to integrate $\int \frac{1}{2}\sin(2x)(3x^{2}-5) \,dx$ . This requires another integration by parts.
Combine all parts of the integral: Let $u = (3x^{2}-5)$ , then $du = 6x \, dx$ . Let $dv = (\frac{1}{2})\sin(2x) \, dx$ , then $v = -(\frac{1}{4})\cos(2x)$ using the integral of sine.
Evaluate the definite integral: Applying integration by parts again, we get $\int (\frac{1}{2})\sin(2x)(3x^{2}-5) \, dx = -(\frac{1}{4})(3x^{2}-5)\cos(2x) - \int -(\frac{1}{4})\cos(2x)(6x) \, dx$ .
Calculate $f(6)$ : The remaining integral is $\int -(\frac{1}{4})\cos(2x)(6x) \, dx$ , which simplifies to $\int -(\frac{3}{2})x \cos(2x) \, dx$ . This requires another integration by parts.
Calculate $f(6)$ : The remaining integral is $\int -(\frac{1}{4})\cos(2x)(6x) \, dx$ , which simplifies to $\int -(\frac{3}{2})x \cos(2x) \, dx$ . This requires another integration by parts.Let $u = -3x$ , then $du = -3 \, dx$ . Let $dv = (\frac{1}{2})\cos(2x) \, dx$ , then $v = (\frac{1}{4})\sin(2x)$ .
Calculate $f(6)$ : The remaining integral is $\int -(\frac{1}{4})\cos(2x)(6x) \, dx$ , which simplifies to $\int -(\frac{3}{2})x \cos(2x) \, dx$ . This requires another integration by parts.Let $u = -3x$ , then $du = -3 \, dx$ . Let $dv = (\frac{1}{2})\cos(2x) \, dx$ , then $v = (\frac{1}{4})\sin(2x)$ .Applying integration by parts, we get $\int -(\frac{3}{2})x \cos(2x) \, dx = (\frac{1}{4})(-3x)\sin(2x) - \int(\frac{1}{4})(-3)\sin(2x) \, dx$ .
Calculate $f(6)$ : The remaining integral is $\int -(\frac{1}{4})\cos(2x)(6x) \, dx$ , which simplifies to $\int -(\frac{3}{2})x \cos(2x) \, dx$ . This requires another integration by parts. Let $u = -3x$ , then $du = -3 \, dx$ . Let $dv = (\frac{1}{2})\cos(2x) \, dx$ , then $v = (\frac{1}{4})\sin(2x)$ . Applying integration by parts, we get $\int -(\frac{3}{2})x \cos(2x) \, dx = (\frac{1}{4})(-3x)\sin(2x) - \int(\frac{1}{4})(-3)\sin(2x) \, dx$ . The remaining integral is $\int(\frac{1}{4})(-3)\sin(2x) \, dx$ , which simplifies to $-(\frac{3}{4}) \int\sin(2x) \, dx$ . The integral of $\int -(\frac{1}{4})\cos(2x)(6x) \, dx$ $0$ is $\int -(\frac{1}{4})\cos(2x)(6x) \, dx$ $1$ .
Calculate $f(6)$ : The remaining integral is $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ , which simplifies to $\int-(\frac{3}{2})x \cos(2x) \, dx$ . This requires another integration by parts.Let $u = -3x$ , then $du = -3 \, dx$ . Let $dv = (\frac{1}{2})\cos(2x) \, dx$ , then $v = (\frac{1}{4})\sin(2x)$ .Applying integration by parts, we get $\int-(\frac{3}{2})x \cos(2x) \, dx = (\frac{1}{4})(-3x)\sin(2x) - \int(\frac{1}{4})(-3)\sin(2x) \, dx$ .The remaining integral is $\int(\frac{1}{4})(-3)\sin(2x) \, dx$ , which simplifies to $-(\frac{3}{4}) \int\sin(2x) \, dx$ . The integral of $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $0$ is $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $1$ .So, $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $2$ . Now we combine all parts of the integral we've calculated.
Calculate $f(6)$ : The remaining integral is $\int -(\frac{1}{4})\cos(2x)(6x) \, dx$ , which simplifies to $\int -(\frac{3}{2})x \cos(2x) \, dx$ . This requires another integration by parts.Let $u = -3x$ , then $du = -3 \, dx$ . Let $dv = (\frac{1}{2})\cos(2x) \, dx$ , then $v = (\frac{1}{4})\sin(2x)$ .Applying integration by parts, we get $\int -(\frac{3}{2})x \cos(2x) \, dx = (\frac{1}{4})(-3x)\sin(2x) - \int(\frac{1}{4})(-3)\sin(2x) \, dx$ .The remaining integral is $\int(\frac{1}{4})(-3)\sin(2x) \, dx$ , which simplifies to $-(\frac{3}{4}) \int\sin(2x) \, dx$ . The integral of $\int -(\frac{1}{4})\cos(2x)(6x) \, dx$ $0$ is $\int -(\frac{1}{4})\cos(2x)(6x) \, dx$ $1$ .So, $\int -(\frac{1}{4})\cos(2x)(6x) \, dx$ $2$ . Now we combine all parts of the integral we've calculated.The integral from $\int -(\frac{1}{4})\cos(2x)(6x) \, dx$ $3$ to $\int -(\frac{1}{4})\cos(2x)(6x) \, dx$ $4$ of $\int -(\frac{1}{4})\cos(2x)(6x) \, dx$ $5$ is the sum of all the parts we've integrated by parts. We need to evaluate this definite integral and add it to $\int -(\frac{1}{4})\cos(2x)(6x) \, dx$ $6$ .
Calculate $f(6)$ : The remaining integral is $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ , which simplifies to $\int-(\frac{3}{2})x \cos(2x) \, dx$ . This requires another integration by parts.Let $u = -3x$ , then $du = -3 \, dx$ . Let $dv = (\frac{1}{2})\cos(2x) \, dx$ , then $v = (\frac{1}{4})\sin(2x)$ .Applying integration by parts, we get $\int-(\frac{3}{2})x \cos(2x) \, dx = (\frac{1}{4})(-3x)\sin(2x) - \int(\frac{1}{4})(-3)\sin(2x) \, dx$ .The remaining integral is $\int(\frac{1}{4})(-3)\sin(2x) \, dx$ , which simplifies to $-(\frac{3}{4}) \int\sin(2x) \, dx$ . The integral of $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $0$ is $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $1$ .So, $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $2$ . Now we combine all parts of the integral we've calculated.The integral from $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $3$ to $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $4$ of $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $5$ is the sum of all the parts we've integrated by parts. We need to evaluate this definite integral and add it to $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $6$ .We evaluate the integral at the bounds $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $4$ and $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $3$ , subtract the lower bound evaluation from the upper bound evaluation, and add the result to $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $6$ .
Calculate $f(6)$ : The remaining integral is $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ , which simplifies to $\int-(\frac{3}{2})x \cos(2x) \, dx$ . This requires another integration by parts.Let $u = -3x$ , then $du = -3 \, dx$ . Let $dv = (\frac{1}{2})\cos(2x) \, dx$ , then $v = (\frac{1}{4})\sin(2x)$ .Applying integration by parts, we get $\int-(\frac{3}{2})x \cos(2x) \, dx = (\frac{1}{4})(-3x)\sin(2x) - \int(\frac{1}{4})(-3)\sin(2x) \, dx$ .The remaining integral is $\int(\frac{1}{4})(-3)\sin(2x) \, dx$ , which simplifies to $-(\frac{3}{4}) \int\sin(2x) \, dx$ . The integral of $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $0$ is $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $1$ .So, $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $2$ . Now we combine all parts of the integral we've calculated.The integral from $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $3$ to $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $4$ of $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $5$ is the sum of all the parts we've integrated by parts. We need to evaluate this definite integral and add it to $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $6$ .We evaluate the integral at the bounds $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $4$ and $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $3$ , subtract the lower bound evaluation from the upper bound evaluation, and add the result to $\int-(\frac{1}{4})\cos(2x)(6x) \, dx$ $6$ .Using a calculator to evaluate the integral at $\int-(\frac{3}{2})x \cos(2x) \, dx$ $0$ and $\int-(\frac{3}{2})x \cos(2x) \, dx$ $1$ , we find the change in $\int-(\frac{3}{2})x \cos(2x) \, dx$ $2$ over this interval. We then add this change to $\int-(\frac{3}{2})x \cos(2x) \, dx$ $3$ to find $f(6)$ .
Calculate $f(6)$ : The remaining integral is $\int -(1/4)\cos(2x)(6x) \, dx$ , which simplifies to $\int -(3/2)x \cos(2x) \, dx$ . This requires another integration by parts. Let $u = -3x$ , then $du = -3 \, dx$ . Let $dv = (1/2)\cos(2x) \, dx$ , then $v = (1/4)\sin(2x)$ . Applying integration by parts, we get $\int -(3/2)x \cos(2x) \, dx = (1/4)(-3x)\sin(2x) - \int (1/4)(-3)\sin(2x) \, dx$ . The remaining integral is $\int (1/4)(-3)\sin(2x) \, dx$ , which simplifies to $-(3/4) \int \sin(2x) \, dx$ . The integral of $\int -(1/4)\cos(2x)(6x) \, dx$ $0$ is $\int -(1/4)\cos(2x)(6x) \, dx$ $1$ . So, $\int -(1/4)\cos(2x)(6x) \, dx$ $2$ . Now we combine all parts of the integral we've calculated. The integral from $\int -(1/4)\cos(2x)(6x) \, dx$ $3$ to $\int -(1/4)\cos(2x)(6x) \, dx$ $4$ of $\int -(1/4)\cos(2x)(6x) \, dx$ $5$ is the sum of all the parts we've integrated by parts. We need to evaluate this definite integral and add it to $\int -(1/4)\cos(2x)(6x) \, dx$ $6$ . We evaluate the integral at the bounds $\int -(1/4)\cos(2x)(6x) \, dx$ $4$ and $\int -(1/4)\cos(2x)(6x) \, dx$ $3$ , subtract the lower bound evaluation from the upper bound evaluation, and add the result to $\int -(1/4)\cos(2x)(6x) \, dx$ $6$ . Using a calculator to evaluate the integral at $\int -(3/2)x \cos(2x) \, dx$ $0$ and $\int -(3/2)x \cos(2x) \, dx$ $1$ , we find the change in $\int -(3/2)x \cos(2x) \, dx$ $2$ over this interval. We then add this change to $\int -(3/2)x \cos(2x) \, dx$ $3$ to find $f(6)$ . After calculating the definite integral and adding it to $\int -(3/2)x \cos(2x) \, dx$ $3$ , we round the result to the nearest thousandth to get the final answer.

The derivative of the function $f$ is defined by $f^{\prime}(x)=\left(x^{3}-5 x\right) \cos (2 x)$ . If $f(-1)=-4$ , then use a calculator to find the value of $f(6)$ to the nearest thousandth. $\newline$ Answer:

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The derivative of the function f f f is defined by f′(x)=(x3−5x)cos⁡(2x) f^{\prime}(x)=\left(x^{3}-5 x\right) \cos (2 x) f′(x)=(x3−5x)cos(2x). If f(−1)=−4 f(-1)=-4 f(−1)=−4, then use a calculator to find the value of f(6) f(6) f(6) to the nearest thousandth.\newlineAnswer:

The derivative of the function $f$ is defined by $f^{\prime}(x)=\left(x^{3}-5 x\right) \cos (2 x)$ . If $f(-1)=-4$ , then use a calculator to find the value of $f(6)$ to the nearest thousandth. $\newline$ Answer: