A curve has a gradient function defined by \frac{dy}{dx} = 6x^2 + 8\sin(4x) + 3. If the cu… — JAMB UTME

A curve has a gradient function defined by $\frac{dy}{dx} = 6x^2 + 8\sin(4x) + 3$ . If the curve passes through the point $(0, 10)$ , what is the value of the constant of integration $C$ ?

Answer: 12

Answer

The constant of integration

C

12

Integrating

6x^2 + 8\sin(4x) + 3

with respect to

x

gives

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

. Substituting

(0, 10)

into the integrated equation gives

10 = 2(0) - 2(1) + 3(0) + C

, which leads directly to

10 = -2 + C

, so

C = 12

Step-by-Step Solution

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

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

The antiderivative of

\sin(kx)

-\frac{1}{k}\cos(kx)

and the antiderivative of

x^n

\frac{x^{n+1}}{n+1}

Apply the given initial condition

(x, y) = (0, 10)

to solve for

C

10 = 2(0)^3 - 2\cos(0) + 3(0) + C \implies 10 = -2 + C

Evaluating at

x = 0

requires evaluating

\cos(0) = 1

, which leaves

-2

from the trigonometric term.

Solve the linear equation for

C

C = 12

Adding

2

to both sides of

10 = -2 + C

gives

C = 12

Key Concept

Indefinite Integration with Boundary Conditions

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