IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

rahul21

^if

_{[0 ]}

^{[1 ]} f(x) dx = 1, _{[0 ]}

^{[ 1]} xf(dx) = a & _{[0 ]}

^{[ 1]} x² f(x) dx a², then _{[0 ]}

^{[ 1]} (a -x)²f(x) dx is equal to ??

edison

_{SINCE, 0 ]}^{[1 ]} f(x) dx = 1, _{[0 ]}

^{[ 1]} xf(dx) = a & _{[0 ]}

^{[ 1]} x² f(x) dx a²,

therefore, I = _{[0 ]}

^{[ 1]} (a -x)²f(x) dx is = _{[0 ]}^{[ 1]} (a²- 2ax + x²)f(x) dx

or I = _{[0 ]}^{[ 1]}a²f(x) dx -2a_{[0 ]}^{[ 1]} x f(x) dx + _{[0 ]}^{[ 1]} x²f(x) dx

or I = a²_{[0 ]}^{[ 1]} f(x) dx - 2a(a) + a^{2 =}a².1 - 2a.a + a²

so, I = 0