CONCEPT:

• \(\smallint \sin x\;dx = \; - \cos x + C\)
• \(\smallint {e^x}\;dx = {e^x} + C\)
• \(\smallint \left[ {f\left( x \right) \pm g\left( x \right)} \right]dx = \;\smallint f\left( x \right)\;dx \pm \;\smallint g\left( x \right)\;dx\)

CALCULATION:

Given:\(\smallint \left( {\sin 2x - 4{e^{3x}}} \right)\;dx = - \frac{1}{a}\cos 2x - \frac{b}{3}\;{e^{3x}} + C\)

As we know that,\(\smallint \left[ {f\left( x \right) \pm g\left( x \right)} \right]dx = \;\smallint f\left( x \right)\;dx \pm \;\smallint g\left( x \right)\;dx\)

⇒\(\smallint \left( {\sin 2x - 4{e^{3x}}} \right)\;dx = \smallint \sin 2x\;dx - \;\smallint 4{e^{3x}}\;dx\)

As we know that,\(\smallint \sin x\;dx = \; - \cos x + C\)and\(\smallint {e^x}\;dx = {e^x} + C\)

⇒\(\smallint \sin 2x\;dx + \;\smallint 4{e^{3x}}\;dx = - \frac{1}{2} cos \ 2x - \frac{4}{3} e^{3x} + C\)

⇒\(\smallint \left( {\sin 2x - 4{e^{3x}}} \right)\;dx = - \frac{1}{2} cos \ 2x - \frac{4}{3} e^{3x} + C\)

Now, by comparing the above equation with\(\smallint \left( {\sin 2x - 4{e^{3x}}} \right)\;dx = - \frac{1}{a}\cos 2x - \frac{b}{3}\;{e^{3x}} + C\)we get,

⇒ a = 2 and b = 4

⇒ a + b = 6

Hence, correct option is 1.