First slide

Evaluation of definite integrals

Question

Let $f$ be a periodic continuous function with period
$T > 0 .$ If $I = \int_{0}^{T} f (x) d x$ Then the value of $I_{1} = \int_{4}^{4 + 4 T} f (3 x) d x$ is

Moderate

A

$I$
B

$2 I$
C

$3 I$
D

$4 I$

Solution

Put $3 x = y$ in $I_{1}$
$\begin{array}{l} I_{1} = \frac{1}{3} \int_{12}^{12 + 12 T} f (y) d y \\ = \frac{1}{3} [\int_{12}^{T} f (y) d y + \sum_{K = 1}^{11} \int_{K T}^{(K + 1) T} f (y) d y + \int_{12 T}^{12 + 12 T} f (y) d y] \end{array}$
But $\int_{K T}^{(K + 1) T} f (y) d y = \int_{0}^{T} f (u) d u,$ (Put $K T + u = y$ or use Property 17)
and $\int_{12 T}^{12 + 12 T} f (y) d y = \int_{0}^{12} f (u) d u$
Hence $I_{1} = \frac{1}{3} [\int_{12}^{T} f (y) d y + 11 I + \int_{0}^{12} f (u) d u]$
$= \frac{1}{3} [\int_{0}^{T} f (y) d y + 11 I] = \frac{1}{3} \times 12 I = 4 I .$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App