Question

Difficulty: HardIndefinite Integration of Polynomial and Trigonometric Functions

A curve y=F(x)y = F(x) has a gradient function given by dydx=9x28x+6cos(3x)4sin(2x)\frac{dy}{dx} = 9x^2 - 8x + 6\cos(3x) - 4\sin(2x). Given that y(0)=7y(0) = 7, determine the value of the constant of integration CC when the antiderivative is expressed in the standard form y=3x34x2+2sin(3x)+2cos(2x)+Cy = 3x^3 - 4x^2 + 2\sin(3x) + 2\cos(2x) + C.

Answer: 5

Answer

The value of the constant of integration CC is 5.
Integrating dydx=9x28x+6cos(3x)4sin(2x)\frac{dy}{dx} = 9x^2 - 8x + 6\cos(3x) - 4\sin(2x) yields y=3x34x2+2sin(3x)+2cos(2x)+Cy = 3x^3 - 4x^2 + 2\sin(3x) + 2\cos(2x) + C. Substituting x=0x = 0 into the expression gives y(0)=00+0+2(1)+C=2+Cy(0) = 0 - 0 + 0 + 2(1) + C = 2 + C. Equating to y(0)=7y(0) = 7 gives 2+C=72 + C = 7, which solves to C=5C = 5.

Step-by-Step Solution

1
Integrate the gradient function term-by-term with respect to xx
y=3x34x2+2sin(3x)+2cos(2x)+Cy = 3x^3 - 4x^2 + 2\sin(3x) + 2\cos(2x) + C
The antiderivative of 9x29x^2 is 3x33x^3, of 8x-8x is 4x2-4x^2, of 6cos(3x)6\cos(3x) is 2sin(3x)2\sin(3x), and of 4sin(2x)-4\sin(2x) is 2cos(2x)2\cos(2x).
2
Apply the initial boundary condition y(0)=7y(0) = 7
3(0)34(0)2+2sin(0)+2cos(0)+C=7    2+C=73(0)^3 - 4(0)^2 + 2\sin(0) + 2\cos(0) + C = 7 \implies 2 + C = 7
At x=0x = 0, sin(0)=0\sin(0) = 0 and cos(0)=1\cos(0) = 1, making the non-zero constant contribution equal to 2(1)=22(1) = 2.
3
Solve for the constant of integration CC
C=5C = 5
Subtracting 2 from both sides of 2+C=72 + C = 7 yields C=5C = 5.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions with Boundary Conditions
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