First slide

Evaluation of definite integrals

Question

lf f(x) is continuous for all real values of x, then $\sum_{n = 1}^{10} \int_{0}^{1} f (r - 1 + x) dx$

Moderate

A

$\int_{0}^{10} f (x) dx$
B

$\int_{0}^{1} f (x) dx$
C

$10 \int_{0}^{1} f (x) dx$
D

$9 \int_{0}^{1} f (x) dx$

Solution

$\begin{matrix} I = \sum_{r = 1}^{10} \int_{0}^{1} f (r - 1 + x) dx \\ I_{t} = \int_{0}^{1} f (t - 1 + x) dx \end{matrix}$
$\begin{array}{l} Put t - 1 + x = y \Rightarrow dx = dy \\ \begin{array}{ll} I_{t} = \int_{t - 1}^{t} f (y) dy \Rightarrow & I_{t} = \int_{t - 1}^{t} f (x) dx \\ I_{1} = \int_{0}^{1} f (x) dx, I_{2} = \int_{1}^{2} f (x) dx \\ I_{3} = \int_{2}^{3} f (x) {dx}_{r . . .} & I_{10} = \int_{9}^{10} f (x) dx \end{array} \end{array} I = I_{1} + I_{2} + . . . . + I_{10} = \int_{0}^{1} f (x) dx + \int_{1}^{2} f (x) dx + . . . . . . . . + \int_{9}^{10} f (x) dx = \int_{0}^{10} f (x) dx$

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