First slide

Evaluation of definite integrals

Question

Let $f$ be a positive function and $I_{1} = \int_{1 - k}^{k} x f (x (1 - x)) d x, I_{2} = \int_{1 - k}^{k} f (x (1 - x)) d x$ where $2 k - 1 > 0$ then $I_{1} / I_{2}$ is

Moderate

A

2
B

k
C

1/2
D

1

Solution

Since $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ we have
$\begin{array}{l} I_{1} = \int_{1 - k}^{k} (k + 1 - k - x) f ((k + 1 - k - x) (1 - (k + 1 - k - x)) \\ = \int_{1 - k}^{k} (1 - x) f ((1 - x) x) d x \\ = \int_{1 - k}^{k} f ((1 - x) x) d x - \int_{1 - k}^{k} x f ((1 - x) x) d x \\ = I_{2} - I_{1} . \end{array}$
So $2 I_{1} = I_{2}$ and $I I_{1} / I_{2} = 1 / 2$ .

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