# How to find integration??

## Given that ${\int}_{0}^{4} h \left(x\right) \mathrm{dx} = 5$, evaluate: a) ${\int}_{0}^{4} 2 h \left(x\right) \mathrm{dx}$ b) ${\int}_{0}^{4} h \left(x\right) - x \mathrm{dx}$ c) ${\int}_{0}^{4} h \left(x\right) + 1 \mathrm{dx}$

Apr 15, 2018

a) 10
b) -3
c) 9

#### Explanation:

a) Remember that a constant can be written outside the integral. Therefore,:
${\int}_{0}^{4} 2 h \left(x\right) \mathrm{dx} = 2 \cdot {\int}_{0}^{4} h \left(x\right) \mathrm{dx} = 2 \cdot 5 = 10$

b) Remember that $\int \left(f \left(x\right) + g \left(x\right)\right) = \int \left(f \left(x\right)\right) + \int \left(g \left(x\right)\right)$

Therefore:
${\int}_{0}^{4} h \left(x\right) - x \mathrm{dx} = {\int}_{0}^{4} h \left(x\right) \mathrm{dx} - {\int}_{0}^{4} x \mathrm{dx}$
${\int}_{0}^{4} x \mathrm{dx} = {\left[\frac{1}{2} {x}^{2}\right]}_{0}^{4} = \frac{16}{2} - \frac{1}{2} \cdot 0 = 8$
${\int}_{0}^{4} h \left(x\right) \mathrm{dx} - {\int}_{0}^{4} x \mathrm{dx} = 5 - 8 = - 3$

c) ${\int}_{0}^{4} h \left(x\right) + 1 \mathrm{dx} = {\int}_{0}^{4} h \left(x\right) \mathrm{dx} + {\int}_{0}^{4} 1 \mathrm{dx}$
${\int}_{0}^{4} 1 \mathrm{dx} = {\left[x\right]}_{0}^{4} = 4 - 0 = 4$
${\int}_{0}^{4} h \left(x\right) \mathrm{dx} + {\int}_{0}^{4} 1 \mathrm{dx} = 5 + 4 = 9$