Calculate the integral by Riemann $ \int _{-1}^{4}( x+1) dx$

Question

Calculate the integral by Riemann $\displaystyle \int _{-1}^{4}( x+1) dx$.

We will choose equal parts: Distance will be $ \Delta x=\frac{4-( -1)}{n}=\frac{5}{n},$ with the left point $a_{k} =-1+ \dfrac{5(k-1)}{n} =-1+\dfrac{5(k-1)}{n}.$

$\begin{array}{ c c l } \lim _{n\rightarrow \infty }\sum _{k=1}^{n}\left( 4+\left( -1+\frac{5( k-1)}{n}\right)\right) \cdotp \frac{5}{n} & = & \lim _{n\rightarrow \infty }\sum _{k=1}^{n}\left( 3+\frac{5( k-1)}{n}\right) \cdotp \frac{5}{n}\\ & = & \lim _{n\rightarrow \infty }\sum _{k=1}^{n}\left(\frac{15}{n} +\frac{25( k-1)}{n^{2}}\right)\\ & = & \lim _{n\rightarrow \infty }\sum _{k=1}^{n}\left(\frac{15}{n}\right) +\sum _{k=1}^{n}\left(\frac{25( k-1)}{n^{2}}\right)\\ & = & \lim _{n\rightarrow \infty }\left(\frac{15}{n}\right) \cdotp n+\frac{25}{n^{2}} \cdotp \sum _{k=1}^{n}( k-1)\\ & = & \lim _{n\rightarrow \infty }\left( 15+\frac{25}{n^{2}} \cdotp \frac{n( n-1)}{3}\right)\\ & = & \lim _{n\rightarrow \infty }\left( 15+\frac{25\left( n^{2} -n\right)}{3n^{2}}\right)\\ & = & \lim _{n\rightarrow \infty }\left( 15+\frac{25\left( 1-\frac{1}{n}\right)}{3}\right)\\ & = & 15+\frac{25}{3}. \end{array}$

I know this solution is wrong. I don't know my mistake.

Answer 1

You should have in the Riemann sum $$ \sum f(a_k)\Delta x $$ with $f(x)=x+1$. But you set $f(x)=4+x$.

Answer 2

Recall that the Riemann sum is computed as the limit where $n\to\infty$ of the sum of the areas of the rectangles that aproximate $f$. Hence, if $a<b$: $$\int_a^bf(x)dx=\lim_{n\to\infty}\frac{b-a}{n}\sum_{i=0}^nf\left(a+\frac{b-a}{n}i\right)$$ where $\frac{b-a}{n}$ is the base of the rectangles and $f\left(a+\frac{b-a}{n}i\right)$ is the height of the rectangles. In our case $a=-1,b=4$ and $f=x+1$, so: $$\int_{-1}^4 (x+1)dx=\lim_{n\to\infty}\frac{5}{n}\sum_{i=0}^n\left(-1+\frac{5}{n}i\right)+1$$ I let you finish this calculation.