Question:
The value of the integral $ \int_{0}^{1} $ $ \frac{e^{5log_{e}x}-e^{4log_{e}x}}{e^{log_{e}x^3}-e^{log_{e}x^2}}dx $ is
The value of the integral $ \int_{0}^{1} $ $ \frac{e^{5log_{e}x}-e^{4log_{e}x}}{e^{log_{e}x^3}-e^{log_{e}x^2}}dx $ is
Updated On: Jun 14, 2022
- $ \frac{1}{3} $
- $ 1 $
- $ -\frac{1}{3} $
- $ -1 $
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The Correct Option is A
Solution and Explanation
Let $I =\int\limits_{0}^{1} \frac{e^{5 log_{e} x} -e^{4 log_{e} x}}{e^{log_{e} x^3}-e^{log_{e}x^2}} dx $
$ = \int\limits_{0}^{1} \frac{e^{log_{e}^5}-e^{log_{e} x^4}}{e^{log_{e} x^3}-e^{log_{e} x^2}} dx $
$ = \int\limits_{0}^{1} \frac{x^{5}-x^{4}}{x^{3}-x^{2}} dx \left[\because e^{log_{e} x} = X\right]$
$ = \int\limits_{0}^{1} \frac{x^{4}\left(x-1\right)}{x^{2}\left(x-1\right)}dx$
$ = \int\limits_{0}^{1} x^{2} dx$
$ = \left[\frac{x^{3}}{3}\right]_{0}^{1} = \frac{1^{3}}{3} - \frac{0^{3}}{3}$
$ \Rightarrow I = \frac{1}{3}$
$ = \int\limits_{0}^{1} \frac{e^{log_{e}^5}-e^{log_{e} x^4}}{e^{log_{e} x^3}-e^{log_{e} x^2}} dx $
$ = \int\limits_{0}^{1} \frac{x^{5}-x^{4}}{x^{3}-x^{2}} dx \left[\because e^{log_{e} x} = X\right]$
$ = \int\limits_{0}^{1} \frac{x^{4}\left(x-1\right)}{x^{2}\left(x-1\right)}dx$
$ = \int\limits_{0}^{1} x^{2} dx$
$ = \left[\frac{x^{3}}{3}\right]_{0}^{1} = \frac{1^{3}}{3} - \frac{0^{3}}{3}$
$ \Rightarrow I = \frac{1}{3}$
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Concepts Used:
Integrals of Some Particular Functions
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
