Evaluate \int_{1}^{4}(1+x+e^{2x})dx as a limit of sums.

 

 

 

 
 
 
 
 

Answers (1)

\int_{1}^{4}(1+x+e^{2x})dx

We know that,

\int_{a}^{b}f(x)dx=(b-a)\lim_{n\rightarrow \infty }\frac{1}{n}\left ( f(a)+f(a+h)+...+f(a+(n-1)h) \right )Putting a=1, b=4 h=\frac{b-a}{n}=\frac{4-1}{n}

\int_{1}^{4}(1+x)dx=(4-1)\lim_{n\rightarrow \infty }\frac{1}{n}\left ( f(1)+f(1+h)+...+f(1+(n-1)h) \right )

Here

f(x)=1+x

f(1)=1+1=2

f(1+h)=1+h+1= 2+h\Rightarrow 2+\frac{3}{h}

f(1+2h)=1+2h+1= 2+2h\Rightarrow 2+2.\frac{3}{h}

f(1+(n-1)h)=1+1+(n-1)h= 2+(n-1)\frac{3}{n}

\Rightarrow (4-1)\lim_{n\rightarrow \infty }\frac{1}{n}\left [ 2+2+\frac{3}{n}+2+2.\frac{3}{n}+...+2+(n-1)\frac{3}{n} \right ]

\Rightarrow (3)\lim_{n\rightarrow \infty }\frac{1}{n}\left [ 2n+\frac{3}{n}\left [ 1+2+...(n-1) \right ] \right ]

We know that 1+2+...(n-1)\Rightarrow \frac{n(n+1)}{2}-n

\Rightarrow (3)\lim_{n\rightarrow \infty }\frac{1}{n}\left [ 2n+\frac{3}{n}\left [ \frac{n(n+1)}{2}-n\right ] \right ]

\Rightarrow (3)\lim_{n\rightarrow \infty }\frac{1}{n}\left [ 2n+\frac{3}{n}\left [ \frac{n(n+1)-2n}{2}\right ] \right ]

\Rightarrow (3)\lim_{n\rightarrow \infty }\frac{1}{n}\left [ 2n+\frac{3}{2n}\left [ n(n+1)-2n\right ] \right ]

\Rightarrow (3)\lim_{n\rightarrow \infty }\frac{1}{n}\left [ 2n+\frac{3n}{2n}\left [ (n+1)-2\right ] \right ]

\Rightarrow (3)\lim_{n\rightarrow \infty }\frac{1}{n}\left [ 2n+\frac{3}{2}\left [ (n-1)\right ] \right ]

\Rightarrow (3)\lim_{n\rightarrow \infty }\frac{1}{n}\left [ 2+\frac{3}{2}\left [ \left ( 1-\frac{1}{n} \right )\right ] \right ]

\Rightarrow (3)\lim_{n\rightarrow \infty }\frac{1}{n}\left [ 2+\frac{3}{2}(1)-\frac{3}{2}\left (\frac{1}{n} \right )\right ]

=3\left [ 2+3/2+0 \right ]\Rightarrow 3\left ( \frac{4+3}{2} \right )=\frac{3}{2}\left ( 7 \right )=\frac{21}{2}

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