# Not sure how to appraoch this ? can someone help would really appreciate it

## Jun 20, 2018

${\int}_{0}^{2} f \left(x\right) \mathrm{dx} = 23$

#### Explanation:

${\int}_{0}^{2} f \left(x\right) \mathrm{dx} = {\int}_{0}^{1} f \left(x\right) \mathrm{dx} + {\int}_{1}^{2} f \left(x\right) \mathrm{dx}$

• ${\int}_{0}^{1} f \left(x\right) \mathrm{dx} = {\int}_{0}^{1} 5 {x}^{9} \mathrm{dx} = 5 {\int}_{0}^{1} \left({x}^{10} / 10\right) ' \mathrm{dx} =$

$5 {\left[{x}^{10} / 10\right]}_{0}^{1} = \frac{5}{10} {\left[{x}^{10}\right]}_{0}^{1} = \frac{1}{2} {\left[{x}^{10}\right]}_{0}^{1} = \frac{1}{2} \left(1 - 0\right) = \frac{1}{2}$

• ${\int}_{1}^{2} f \left(x\right) \mathrm{dx} = {\int}_{1}^{2} 6 {x}^{3} \mathrm{dx} = 6 {\int}_{1}^{2} \left({x}^{4} / 4\right) ' \mathrm{dx} =$

$\frac{6}{4} {\int}_{1}^{2} \left({x}^{4}\right) ' \mathrm{dx} = \frac{3}{2} {\left[{x}^{4}\right]}_{1}^{2} = \frac{3}{2} \left(16 - 1\right) = \frac{3}{2} \cdot 15 = \frac{45}{2}$

Hence, ${\int}_{0}^{2} f \left(x\right) \mathrm{dx} = \frac{1}{2} + \frac{45}{2} = \frac{46}{2} = 23$