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प्रश्न
Evaluate: `int_0^3 f(x)dx` where f(x) = `{(cos 2x, 0<= x <= pi/2),(3, pi/2 <= x <= 3) :}`
उत्तर
`int_0^3 f(x) dx = int_0^(pi/2) cos 2x dx + int_(pi/2)^3 dx`
`= (sin 2x/2)_0^(pi/2) + [3x]_(pi/2)^3`
`= [(sin 2 xx pi/2)/2 - (sin 2 xx 0)/2] + [(3 xx 3 - 3 xx pi/2)]`
`= 0 + 9 - (3pi)/2`
`= 9- (3pi)/2`
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