MCQ
$\int_0^1 {f(1 - x)\,dx} $ has the same value as the integral
- ✓$\int_0^1 {f(x)\,dx} $
- B$\int_0^1 {f( - x)\,dx} $
- C$\int_0^1 {f(x - 1)\,dx} $
- D$\int_{ - 1}^1 {f(x)\,dx} $
Also as $x = 0$ to $1,$ $t = 1$ to $0$
Therefore, $\int_0^1 {f(1 - x)dx = \int_1^0 {f(t)( - dt)} } = \int_0^1 {f(t)dt = \int_0^1 {f(x)dx} } $.
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