Question

please help with this questions with steps

Answer 1

$\textsf{\large{\underline{Solution}:}}$

We have to evaluate the given integral.

$\displaystyle = \sf\int_{1}^{e} \: \dfrac{1}{x} \: dx$

We know that:

$\displaystyle: \longmapsto \sf\int \: \dfrac{1}{x} \: dx = ln(x) + C$

Therefore, we get:

$\displaystyle = \sf ln(x) \bigg|_{1}^{e}$

$\displaystyle = \sf ln(e) - ln(1)$

As we know - e¹ = e and e⁰ = 1. We get:

$\displaystyle = \sf1 - 0$

$= \sf1$

Therefore:

$\displaystyle: \longmapsto \sf\int_{1}^{e} \: \dfrac{1}{x} \: dx = 1$

$\textsf{\large{\underline{Answer}:}}$

• The result obtained after evaluating the given integral is 1.

$\textsf{\large{\underline{Additional Information}:}}$

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf kx+C\\ \\ \sf sin(x)&\sf-cos(x)+C\\ \\ \sf cos(x)&\sf sin(x)+C\\ \\ \sf{sec}^{2}(x)&\sf tan(x)+C\\ \\ \sf{cosec}^{2}(x)&\sf-cot(x)+C\\ \\ \sf sec(x)\ tan(x)&\sf sec(x)+C\\ \\ \sf cosec(x)\ cot(x)&\sf-cosec(x)+C\\ \\ \sf tan(x)&\sf log(sec(x))+C\\ \\ \sf\dfrac{1}{x}&\sf ln(x)+C\\ \\ \sf{e}^{x}&\sf{e}^{x}+C\\ \\ \sf x^{n},n\neq-1&\sf\dfrac{x^{n+1}}{n+1}+C\end{array}}$