Question

Evaluate the following integral by expressing them as a limit of a sum.
${\int }_1^2(3x-2)dx$

• A. $\frac{1}{2}$
• B. $\frac{3}{2}$
• C. $\frac{5}{2}$
• D. $\frac{7}{2}$

Answer 1

The correct option is C $\frac{5}{2}$

${\int }_1^2(3x-2)dx{\int }_a^{b}f\left(x\right).dx=\underset{h\to 0}{lim}h\left[f\right(a)+f(a+h)+f(a+2h)+....+f(a+(n-1)h]$

where $h=\frac{b-a}{n}$

$\therefore h=\frac{2-1}{n}=\frac{1}{n}$

$I={\int }_1^2(3x-2).dx=\underset{h\to 0}{lim}h\left[f\right(1)+f(1+h)+f(1+2h)+...+f(1+(n-1)h\left)\right]$

$=\underset{h\to 0}{lim}h\left[f\right(1)+f[3(1+a)-2]+[3(1+2h)-2]+....+[3(1+(n-1\left)h\right]-2]$

$=\underset{h\to 0}{lim}h[1+3+3h-2+3+6h-2+......+3+3(n-1)h-2]$

$=\underset{h\to 0}{lim}h[3n+3h(1+2+3+....+(n-1)]$

$=\underset{h\to 0}{lim}h[n+3h\times \frac{n(n-1)}{2}]=\underset{h\to \infty }{lim}\frac{1}{n}\{n+3\times \frac{1}{n}\times \frac{n(n-1)}{2}\}$

$=\underset{n\to \infty }{lim}\frac{1}{n}\{n+\frac{3n-3}{2}\}=\underset{n\to \infty }{lim}\frac{1}{n}\left\{\frac{2n+3n-3}{2}\right\}$

$=\underset{n\to \infty }{lim}\frac{1}{n}\left\{\frac{5n-3}{2}\right\}=\underset{n\to \infty }{lim}(\frac{5}{2}-\frac{3}{2n})=\frac{5}{2}$