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Question

f(x) is a cubic polynomial function such that f(x)=0 at x=1,2,3 and passing through the point (0,6). Then the area bounded by the curve f(x) and the x axis between x=0 and x=3 is

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Solution

f(x)=0 at x=1,2,3
f(x)=a(x1)(x2)(x3)
Since, f(x) is passing through the point (0,6)
f(0)=6
a(01)(02)(03)=6
a=1
f(x)=(x1)(x2)(x3)

The graph of the given function for 0x3 is as shown in figure.

Hence, the required area A= shaded area
=∣ ∣10y dx∣ ∣+∣ ∣21y dx∣ ∣+∣ ∣32y dx∣ ∣ ...(1)
Since, y dx=(x1)(x2)(x3)dx
=(x36x2+11x6)dx
=x442x3+11x226x
From (1)
A=∣ ∣[x442x3+11x226x]10∣ ∣+∣ ∣[x442x3+11x226x]21∣ ∣+∣ ∣[x442x3+11x226x]32∣ ∣
=94+(14)+14
=114 sq. units

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