MHT CET202615 April 2026Morning ShiftMathematicsIndefinite IntegrationActual
If f(x) and g(x) are integrable functions then [ f(x)dx ] [ g(x)dx ] =
Options
- A[f(x)g'(x) + f'(x)g(x) ]dx
- B[f(x)g'(x) - f'(x)g(x) ]dx
- C[f(x) g(x)dx + g(x) f(x)dx ]dx
- D[f(x) g(x)dx - g(x) f(x)dx ]dx
Correct answer
C. [f(x) g(x)dx + g(x) f(x)dx ]dx
Step-by-step solution
Let u = f(x)dx and v = g(x)dx . Differentiating with respect to x , we get du dx = f(x) and dv dx = g(x) . Using the product rule of differentiation, d dx (uv) = u dv dx + v du dx . Substituting the expressions for u , v , du dx , and dv dx , we get d dx [ ( f(x)dx ) ( g(x)dx ) ] = ( f(x)dx )g(x) + ( g(x)dx )f(x) . Integrating both sides with respect to x , we obtain [ f(x)dx ] [ g(x)dx ] = [f(x) g(x)dx + g(x) f(x)dx ]dx . Answer: [f(x) g(x)dx + g(x) f(x)dx ]dx