Definite integral: Practice

We can use Riemann sums here, its definition is: $\int_a^bf(x)\,\mathrm dx=\lim\limits_{n\to\infty}\sum_{i=1}^{n}f(x_i^*)\Delta x$ where $\Delta x=\dfrac{b-a}{n}$ and $x_i^*=a+(\Delta x)i$ . For our problem, we recognize that: $\Delta x=\dfrac{3-1}{n}=\dfrac2n$ and so: $x_i^*=1+\dfrac{2i}n$ , therefore: $\begin{aligned} f(x_i^*)&=\left(1+\dfrac{2i}{n}\right)^2-\left(1+\dfrac{2i}{n}\right)+1\\ &=1+\dfrac{4i}{n}+\dfrac{4i^2}{n^2}\cancel{-1}-\dfrac{2i}{n}\cancel{+1}\\ &=1+\dfrac{2i}{n}+\dfrac{4i^2}{n^2}. \end{aligned}$ So by definition: $\begin{aligned} \int_1^3(x^2-x+1)\,\mathrm dx&=\lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x\\ &=\lim_{n\to\infty}\sum_{i=1}^n \left[1+\dfrac{2i}{n}+\dfrac{4i^2}{n^2}\right]\dfrac2n\\ &=\lim_{n\to\infty} \left[\sum_{i=1}^n1+\sum_{i=1}^n\dfrac{2i}{n}+\sum_{i=1}^n\dfrac{4i^2}{n^2}\right]\dfrac2n\\ &=\lim_{n\to\infty} \left[\color{darkmagenta}{\sum_{i=1}^n1}+\dfrac{2}{n}\color{#3D99F6}{\sum_{i=1}^ni}+\dfrac{4}{n^2}\color{#20A900}{\sum_{i=1}^ni^2}\right]\dfrac2n\\ &\overset{\displaystyle(\star)}=\lim_{n\to\infty} \left[\color{darkmagenta}n+\dfrac{2}{n}\color{#3D99F6}{\dfrac{n(n+1)}{2}}+\dfrac{4}{n^2}\color{#20A900}{\dfrac{n(n+1)(2n+1)}{6}}\right]\dfrac2n\\ &=\lim_{n\to\infty} \left[2+\dfrac{4}{n^2}\dfrac{n(n+1)}{2}+\dfrac{8}{n^3}\dfrac{n(n+1)(2n+1)}{6}\right]\\ &=\lim_{n\to\infty} \left[2+2\dfrac{n^2+n}{n^2}+\dfrac{4}{3}\dfrac{2n^3+\mathcal O(n)}{n^3}\right]\\ &=2+2+\dfrac43\cdot2\\ &=4+\dfrac83=\dfrac{20}3. \end{aligned}$

put f(x)=xsquare-x+1 and solve

Use the power rule for antiderivatives. You get: $\left(\left. \frac{x^3}{3}-\frac{x^2}{2}+x \right) \right|_{1}^{3}$

$=\frac{54}{6}-\frac{27}{6}+\frac{18}{6}-\frac{2}{6}+\frac{3}{6}-\frac{6}{6}$

$=\frac{40}{6} = \boxed{\frac{20}{3}}$