If \int (6x^2 - 4\sin(2x)) \, dx = ax^3 + b\cos(2x) + C, where a, b, and C are constants, … — JAMB UTME

If $\int (6x^2 - 4\sin(2x)) \, dx = ax^3 + b\cos(2x) + C$ , where $a$ , $b$ , and $C$ are constants, what is the value of $a + b$ ?

Answer: 4

The value of

a + b

4

Integrating

6x^2

gives

2x^3

, and integrating

-4\sin(2x)

gives

2\cos(2x)

. Comparing

2x^3 + 2\cos(2x) + C

ax^3 + b\cos(2x) + C

yields

a=2

and

b=2

, giving

a+b=4

Integrate each term of the integrand

(6x^2 - 4\sin(2x))

with respect to

x

\int 6x^2 \, dx = \frac{6x^3}{3} = 2x^3

, and

\int -4\sin(2x) \, dx = -4 \cdot \left(-\frac{1}{2}\cos(2x)\right) = 2\cos(2x)

Applying the power rule for integration

\int x^n \, dx = \frac{x^{n+1}}{n+1}

and the standard trigonometric integral formula

\int \sin(kx) \, dx = -\frac{1}{k}\cos(kx)

Combine the calculated antiderivative terms and include the constant of integration

C

\int (6x^2 - 4\sin(2x)) \, dx = 2x^3 + 2\cos(2x) + C

Summing the individual term-by-term antiderivatives produces the complete indefinite integral.

Compare the resulting expression with

ax^3 + b\cos(2x) + C

to determine

a

and

b

a = 2

and

b = 2

Matching corresponding coefficients of

x^3

and

\cos(2x)

Compute

a + b

2 + 2 = 4

Evaluating the sum of the extracted coefficients.

Indefinite Integration of Polynomial and Trigonometric Functions

Estimated Time:1m 30s

Question