First slide

Evaluation of definite integrals

Question

Let $p (x)$ be a function defined on R such that $p' (x) = p' (1 - x)$ , for all $x \in [0, 1], p (0) = 1$ and $p (1) = 41 .$ Then $\int_{0}^{1} p (x) d x$ equals

Moderate

A

41
B

42
C

$\sqrt{41}$
D

21

Solution

$\int_{0}^{1} p (x) d x = \int_{0}^{1} 1 \cdot p (x) d x = [x p (x)]_{0}^{1} - I_{1} = p (1) - I_{1}$
where
$\begin{array}{l} I_{1} = \int_{0}^{1} x p^{'} (x) d x = \int_{0}^{1} (1 - x) p^{'} (1 - x) d x \\ = \int_{0}^{1} (1 - x) p^{'} (x) d x = \int_{0}^{1} p^{'} (x) d x - I_{1} \end{array}$
$\Rightarrow 2 I_{1} = [p (x)]_{0}^{1} = p (1) - p (0)$
Thus,
$\begin{aligned} \int_{0}^{1} p (x) d x = p (1) - \frac{1}{2} (p (1) - p (0)) \\ = \frac{1}{2} (p (1) + p (0)) = \frac{1}{2} (41 + 1) = 21 \end{aligned}$

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