First slide

Evaluation of definite integrals

Question

$\int_{0}^{6} x (x - 1) (x - 2) (x - 3) (x - 4) (x - 5) (x - 6) d x$

Moderate

A

0
B

1
C

4
D

5

Solution

$The point of symmetry is (3, 0)$
$\begin{array}{l} We make the change of variable x = t + 3 \\ Then g (t) = (t + 3) (t + 2) (t + 1) t (t - 1) (t - 2) (t - 3) \end{array}$
$\begin{aligned} ∴ \int_{0}^{6} x (x - 1) (x - 2) (x - 3) (x - 4) (x - 5) (x - 6) d x \\ = \int_{- 3}^{3} g (t) d t but g (- t) = - g (t) \end{aligned}$
$t h e r e f o r e g (t) is odd function$
$= 0$

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