First slide

Introduction to integration

Question

$If \int \frac{2^{x + 1} - 5^{x - 1}}{10^{x}} d x = = A \frac{{(\frac{1}{5})}^{x}}{\log (\frac{1}{5})} + B \frac{{(\frac{1}{2})}^{x}}{\log (\frac{1}{2})} + c t h e n$

Easy

A

$A = \frac{2}{5}, B = \frac{1}{5}$
B

$A = \frac{2}{5}, B = \frac{- 1}{5}$
C

$A = 2, B = - \frac{1}{5}$
D

$A = 2, B = \frac{1}{5}$

Solution

$\begin{array}{l} I = \int \frac{2^{x + 1} - 5^{x - 1}}{10^{x}} d x \\ = \int \frac{5^{x} 2^{x} (\frac{2}{5^{x}} - \frac{1}{5 \cdot 2^{x}})}{10^{x}} d x \\ = \int [2 {(\frac{1}{5})}^{x} - \frac{1}{5} {(\frac{1}{2})}^{x}] d x \\ = \frac{2 {(\frac{1}{5})}^{x}}{\log (\frac{1}{5})} - \frac{1}{5} \frac{{(\frac{1}{2})}^{x}}{\log (\frac{1}{2})} + c f o r m u l a \int a^{x} d x = \frac{a^{x}}{\log_{e} a} + c \end{array}$

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