# What is the area under f(x)=x^2-3x+5 in x in[0,1] ?

Dec 10, 2015

$\frac{23}{6}$ square units

#### Explanation:

Drawing the graph of the function we get

graph{x^2-3x+5 [-9.14, 19.34, -3.53, 10.71]}

Since there are no x-intercepts or turning points between the interval [0,1], we may integrate throughout the interval to obtain the required area as

${\int}_{0}^{1} \left({x}^{2} - 3 x + 5\right) \mathrm{dx}$

$= {\left[{x}^{3} / 3 - 3 {x}^{2} / 2 + 5 x\right]}_{0}^{1}$

$= \frac{1}{3} - \frac{3}{2} + 5$

$= \frac{23}{6}$