Question

simplify and write in exponential form (-41)⁶ ÷ (-41)³​

Answer 1

Answer:

Explanation :-

Given limit can be written as

$\begin{gathered}\displaystyle \lim_{ n \to \infty} \frac{\int_{0}^{ \frac{1}{n} } {x}^{2018x + 1} \: dx}{ \frac{1}{ {n}^{2} } } \\ \end{gathered} </p><p>n→∞</p><p>lim</p><p> </p><p> </p><p>n </p><p>2</p><p> </p><p>1</p><p> </p><p> </p><p> \\ \\ ∫ </p><p>0</p><p>n</p><p>1</p><p> </p><p> </p><p> </p><p> x </p><p>2018x+1</p><p> dx$

Now use L hospital rule . To differentiate numerator we will use fundamental theorem of calculas .

$\begin{gathered} \frac{d}{dx} \int_{0} ^{y} f(t) \: dt = f(y). \frac{dy}{dx} \\ \end{gathered} </p><p>dx</p><p>d</p><p> </p><p> \\ ∫ </p><p>0</p><p>y</p><p> </p><p> f(t)dt=f(y). </p><p>dx</p><p>dy</p><p>$

Our limit becomes

$\begin{gathered}\displaystyle \lim_{ n \to \infty} \: \frac{( \frac{1}{n} ) ^{2018. \frac{1}{n} + 1} . \frac{d}{dn}( \frac{1}{n} )}{ \frac{ - 2}{ {n}^{3} } } \\ \\ = \displaystyle \lim_{ n \to \infty} \: \frac{\bigg( \frac{1}{n} \bigg) ^{2018. \frac{1}{n} } \frac{1}{n} . ( \frac{ - 1}{n {}^{2} } )}{ \frac{ - 2}{ {n}^{3} } } \\ \\ = \displaystyle \lim_{ n \to \infty} \: \frac{\bigg( \frac{1}{n} \bigg) ^{2018. \frac{1}{n}} } {2} \\ \\ = \displaystyle \frac{1}{2} \lim_{ n \to \infty} \exp \bigg\{ \log\bigg( \frac{1}{n} \bigg) ^{2018. \frac{1}{n}} \bigg \} \\ \\ = \frac{1}{2} \lim_{ n \to \infty} \exp \bigg\{ \frac{2018}{n} \log\bigg( \frac{1}{n} \bigg) \bigg \} \\ \\ again \: l \: hopital \: rule \\ \\ = \frac{1}{2} \lim_{ n \to \infty} \exp \bigg\{ 2018\bigg( \frac{1}{ \frac{1}{n} } \bigg) .\frac{ - 1}{ {n}^{2} } \bigg \} \\ \\ = \frac{1}{2} \exp \{0 \} = \frac{1}{2} \end{gathered} </p><p>n→∞</p><p>lim</p><p> </p><p> </p><p>n </p><p>3</p><p> </p><p>−2</p><p> </p><p> </p><p>( </p><p>n</p><p>1</p><p> </p><p> ) </p><p>2018. </p><p>n</p><p>1</p><p> </p><p> +1</p><p> . </p><p>dn</p><p>d</p><p> </p><p> ( </p><p>n</p><p>1</p><p> </p><p> )</p><p> </p><p> </p><p>= </p><p>n→∞</p><p>lim</p><p> </p><p> </p><p>n </p><p>3</p><p> </p><p>−2</p><p> </p><p> </p><p>( </p><p>n</p><p>1</p><p> </p><p> ) </p><p>2018. </p><p>n</p><p>1</p><p> </p><p> </p><p> </p><p>n</p><p>1</p><p> </p><p> .( </p><p>n </p><p>2</p><p> </p><p>−1</p><p> </p><p> )</p><p> </p><p> </p><p>= </p><p>n→∞</p><p>lim</p><p> </p><p> </p><p>2</p><p>( </p><p>n</p><p>1</p><p> </p><p> ) </p><p>2018. </p><p>n</p><p>1</p><p> </p><p> </p><p> </p><p> </p><p> </p><p>= </p><p>2</p><p>1</p><p> </p><p> </p><p>n→∞</p><p>lim</p><p> </p><p> exp{log( </p><p>n</p><p>1</p><p> </p><p> ) </p><p>2018. </p><p>n</p><p>1</p><p> </p><p> </p><p> }</p><p>= </p><p>2</p><p>1</p><p> </p><p> </p><p>n→∞</p><p>lim</p><p> </p><p> exp{ </p><p>n</p><p>2018</p><p> </p><p> log( </p><p>n</p><p>1</p><p> </p><p> )}</p><p>againlhopitalrule</p><p>= </p><p>2</p><p>1</p><p> </p><p> </p><p>n→∞</p><p>lim</p><p> </p><p> exp{2018( </p><p>n</p><p>1</p><p> </p><p> </p><p>1</p><p> </p><p> ). </p><p>n </p><p>2</p><p> </p><p>−1</p><p> </p><p> }</p><p>= </p><p>2</p><p>1</p><p> </p><p> exp{0}= </p><p>2</p><p>1</p><p> </p><p> </p><p> </p><p> </p><p></p><p>$

Answer 2

Answer:

$(-41)^{3}$

Step-by-step explanation:

$\frac{(-41)(-41)(-41)(-41)(-41)(-41)}{(-41)(-41)(-41)}$

=(-41)(-41)(-41)=(-41)³​