The derivative of the function f is defined by f′(x)=(x2−3x)cos(2x). If f(0)=5, then use a calculator to find the value of f(6) to the nearest thousandth.Answer:
Q. The derivative of the function f is defined by f′(x)=(x2−3x)cos(2x). If f(0)=5, then use a calculator to find the value of f(6) to the nearest thousandth.Answer:
Integrate f′(x) for f(x): To find f(6), we need to integrate the derivative f′(x) to get the original function f(x). We will then use the initial condition f(0)=5 to find the constant of integration.
Use Integration by Parts: The integral of f′(x)=(x2−3x)cos(2x) is not straightforward due to the product of a polynomial and a trigonometric function. We will use integration by parts or a computer algebra system to find the integral.
Find du and v: Let's denote u=x2−3x and dv=cos(2x)dx. Then we need to find du and v: du=(2x−3)dx and v=21sin(2x).
Integrate by Parts Again: Applying integration by parts, we get:∫f′(x)dx=∫udv=uv−∫vdu= (x2−3x)(21)sin(2x)−∫(21)sin(2x)(2x−3)dx.
Evaluate Initial Condition: We need to integrate (21)sin(2x)(2x−3)dx, which again requires integration by parts or a computer algebra system.
Calculate Constant of Integration: Let's denote u=2x−3 and dv=(21)sin(2x)dx. Then we need to find du and v: du=2dx and v=−(41)cos(2x).
Final Function f(x): Applying integration by parts again, we get:∫(21)sin(2x)(2x−3)dx=uv−∫vdu= −(41)(2x−3)cos(2x) - \int -(\frac{1}{4})\cos(2x)(2)\,dx.
Evaluate f(6): The integral of −41cos(2x)(2)dx is −41(2)(21)sin(2x)=−4sin(2x).
Evaluate f(6): The integral of −41cos(2x)(2)dx is −41(2)21sin(2x)=−4sin(2x). Combining all parts, we get: f(x)=(x2−3x)21sin(2x)+41(2x−3)cos(2x)+4sin(2x)+C, where C is the constant of integration.
Evaluate f(6): The integral of −41cos(2x)(2)dx is −41(2)21sin(2x)=−4sin(2x). Combining all parts, we get: f(x)=(x2−3x)21sin(2x)+41(2x−3)cos(2x)+4sin(2x)+C, where C is the constant of integration. Using the initial condition f(0)=5, we find C: f(0)=(02−3⋅0)21sin(2⋅0)+41(2⋅0−3)cos(2⋅0)+4sin(2⋅0)+C=5. Since sin(0)=0 and cos(0)=1, this simplifies to: −41cos(2x)(2)dx0.
Evaluate f(6): The integral of −41cos(2x)(2)dx is −41(2)21sin(2x)=−4sin(2x). Combining all parts, we get: f(x)=(x2−3x)21sin(2x)+41(2x−3)cos(2x)+4sin(2x)+C, where C is the constant of integration. Using the initial condition f(0)=5, we find C: f(0)=(02−3⋅0)21sin(2⋅0)+41(2⋅0−3)cos(2⋅0)+4sin(2⋅0)+C=5. Since sin(0)=0 and cos(0)=1, this simplifies to: −41cos(2x)(2)dx0. Now we have the function −41cos(2x)(2)dx1.
Evaluate f(6): The integral of −41cos(2x)(2)dx is −41(2)21sin(2x)=−4sin(2x). Combining all parts, we get: f(x)=(x2−3x)21sin(2x)+41(2x−3)cos(2x)+4sin(2x)+C, where C is the constant of integration. Using the initial condition f(0)=5, we find C: f(0)=(02−3⋅0)21sin(2⋅0)+41(2⋅0−3)cos(2⋅0)+4sin(2⋅0)+C=5. Since sin(0)=0 and cos(0)=1, this simplifies to: C=5+43. Now we have the function f(x)=(x2−3x)21sin(2x)+41(2x−3)cos(2x)+4sin(2x)+5+43. We can now evaluate f(6) using a calculator: f(6)=(62−3⋅6)21sin(2⋅6)+41(2⋅6−3)cos(2⋅6)+4sin(2⋅6)+5+43.
Evaluate f(6): The integral of −41cos(2x)(2)dx is −41(2)21sin(2x)=−4sin(2x). Combining all parts, we get: f(x)=(x2−3x)21sin(2x)+41(2x−3)cos(2x)+4sin(2x)+C, where C is the constant of integration. Using the initial condition f(0)=5, we find C: f(0)=(02−3⋅0)21sin(2⋅0)+41(2⋅0−3)cos(2⋅0)+4sin(2⋅0)+C=5. Since sin(0)=0 and cos(0)=1, this simplifies to: −41cos(2x)(2)dx0. Now we have the function −41cos(2x)(2)dx1. We can now evaluate f(6) using a calculator: −41cos(2x)(2)dx3. After calculating the above expression using a calculator, we find the value of f(6) to the nearest thousandth.
More problems from Find the roots of factored polynomials