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NCERT Class XII Mathematics Chapter - - Solutions

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Question : 226 of 271
Marks: +1, -0
Show that ∫0af(x)g(x)dx=2∫0af(x)dx,\int\limits_{0}^{a} f(x) g(x) dx = 2 \int\limits_{0}^{a} f(x) dx, if ff and gg are defined as f(x)=f(a−x)f(x) = f(a - x) and g(x)+g(a−x)=4g(x) + g(a - x) = 4
Solution:  
Let ∫0af(x)g(x)dx\int\limits_{0}^{a} f(x) g(x) dx =∫0af(a−x)[4−g(a−x)]dx= \int\limits_{0}^{a} f(a-x) [4 - g(a-x)] dx
=4∫0af(a−x)dx−∫0af(a−x)g(a−x)dx= 4 \int\limits_{0}^{a} f(a-x) dx - \int\limits_{0}^{a} f(a-x) g(a-x) dx
Let a−x=t⇒−dx=dta - x = t \Rightarrow -dx = dt
When x=0,t=ax = 0, t = a and when x=a,t=0x = a, t = 0
I=−4∫a0f(t)dt+∫0af(t)g(t)dtI = -4 \int\limits_{a}^{0} f(t) dt + \int\limits_{0}^{a} f(t) g(t) dt
=4∫0af(t)dt−∫0af(t)g(t)dt= 4 \int\limits_{0}^{a} f(t) dt - \int\limits_{0}^{a} f(t) g(t) dt
=4∫0af(x)dx−∫0af(x)g(x)dx= 4 \int\limits_{0}^{a} f(x) dx - \int\limits_{0}^{a} f(x) g(x) dx
=4∫0af(x)dx−I= 4 \int\limits_{0}^{a} f(x) dx - I
⇒2I=4∫0af(x)dx\Rightarrow 2I = 4 \int\limits_{0}^{a} f(x) dx
Hence I=2∫0af(x)dxI = 2 \int\limits_{0}^{a} f(x) dx
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