A curve has a gradient function defined by \frac{dy}{dx} = 12x^3 - 6\sin(3x) + 4. If the c… — JAMB UTME

A curve has a gradient function defined by $\frac{dy}{dx} = 12x^3 - 6\sin(3x) + 4$ . If the curve passes through the point $(0, 15)$ , what is the value of the constant of integration $C$ ?

Answer: 13

Answer

The constant of integration

C

is 13.

Integrating

12x^3 - 6\sin(3x) + 4

with respect to

x

yields

y = 3x^4 + 2\cos(3x) + 4x + C

. Substituting

x = 0

and

y = 15

gives

15 = 3(0)^4 + 2\cos(0) + 4(0) + C = 2 + C

, which simplifies to

C = 13

Step-by-Step Solution

Integrate the gradient function to obtain the general equation of the curve.

y = 3x^4 + 2\cos(3x) + 4x + C

The integral of

12x^3

3x^4

, the integral of

-6\sin(3x)

+2\cos(3x)

, and the integral of

4

4x

Apply the initial condition

(0, 15)

to solve for

C

15 = 3(0)^4 + 2\cos(0) + 4(0) + C

Substitute

x = 0

and

y = 15

into the integrated equation.

Calculate the numerical value of

C

C = 13

Since

\cos(0) = 1

, the equation becomes

15 = 2 + C

, yielding

C = 13

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions with Initial Conditions

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