Question

Evaluate using the definition of the definite integral (that means using the limit of a Riemann sum)

Answer 1

The Riemann Sum Definite Integral Formula,

`\int ^b_af(x)dx=\lim _(n\to\infty)\sum ^n_(i\mathop=1)f(x^{}_(i^(\ast)))\Delta x`

Where

`\begin{gathered} \Delta x=(b-a)/(n) \\ x_(i^(\ast))=a+(\Delta x)i \\ \end{gathered}`

and,

`\begin{gathered} \sum ^n_(i\mathop=1)k=kn \\ \sum ^n_{i\mathop{=}1}i=(n(n+1))/(2) \\ \sum ^n_{i\mathop{=}1}i^2=(n(n+1)(2n+1))/(6) \end{gathered}`

The integral to evaluate is

`\int ^b_ax^2dx`

Lets find x_i *:

`\begin{gathered} x_(i^(\ast))=a+(\Delta x)i \\ x_(i^(\ast))=a+((b-a)/(n))i \\ x_(i^(\ast))=a+((b-a)i)/(n) \\ x_(i^(\ast))=a+(bi-ai)/(n) \\ x_(i^(\ast))=(an+bi-ai)/(n) \end{gathered}`

Now, let's find f(x_i*):

`f(x_(i^(\ast)))=((an+bi-ai)/(n))^2=((an+bi-ai)/(n))((an+bi-ai)/(n))=\frac{a^2n^2+2\text{anbi}-2a^2ni-2\text{abi}^2+a^2i^2+b^2i^2}{n^2}`

Now, putting it into the formula,

`\begin{gathered} \int ^b_af(x)dx=\lim _(n\to\infty)\sum ^n_{i\mathop{=}1}f(x^{}_(i^(\ast)))\Delta x \\ \lim _(n\to\infty)\sum ^n_{i\mathop{=}1}\frac{a^2n^2+2\text{anbi}-2a^2ni-2\text{abi}^2+a^2i^2+b^2i^2}{n^2}*(b-a)/(n) \end{gathered}`

Simplifying:

`\begin{gathered} \lim _(n\to\infty)\sum ^n_{i\mathop{=}1}(\frac{a^2n^2+2\text{anbi}-2a^2ni-2\text{abi}^2+a^2i^2+b^2i^2}{n^2})*(b-a)/(n) \\ \lim _(n\to\infty)\sum ^n_{i\mathop{=}1}(a^2+\frac{2\text{abi}}{n}-(2a^2i)/(n)-\frac{2\text{abi}^2}{n^2}+(a^2i^2)/(n^2)+(b^2i^2)/(n^2))*(b-a)/(n) \\ \lim _(n\to\infty)((b-a)/(n))\lbrack\sum ^n_{i\mathop{=}1}a^2+\sum ^n_{i\mathop{=}1}\frac{2\text{abi}}{n}-\sum ^n_{i\mathop{=}1}(2a^2i)/(n)-\sum ^n_{i\mathop{=}1}\frac{2\text{abi}^2}{n^2}+\sum ^n_{i\mathop{=}1}(a^2i^2)/(n^2)+\sum ^n_{i\mathop{=}1}(b^2i^2)/(n^2)\rbrack \end{gathered}`

Further simplifying: