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

Question

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

Emilia Aguilar · Accepted Answer

#23/6# square units

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(x^2-3x+5)dx# #=[x^3/3-3x^2/2+5x]_0^1# #=1/3-3/2+5# #=23/6#

Aurora Alvarado · Answer

undefined To find the area under the curve $f(x) = x^2 - 3x + 5$ in the interval $[0, 1]$, you need to compute the definite integral of $f(x)$ with respect to $x$ over that interval. Using the fundamental theorem of calculus, integrate $f(x)$ from 0 to 1:

$$
\int_{0}^{1} (x^2 - 3x + 5) \, dx
$$

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

$$= \left(\frac{1}{3} - \frac{3}{2} + 5\right) - \left(0 - 0 + 0\right)$$

$$= \left(\frac{1}{3} - \frac{3}{2} + 5\right)$$

$$= \left(\frac{1}{3} - \frac{9}{6} + \frac{30}{6}\right)$$

$$= \left(\frac{1}{3} + \frac{21}{6}\right)$$

$$= \left(\frac{1}{3} + \frac{7}{2}\right)$$

$$= \left(\frac{1}{3} + \frac{21}{6}\right)$$

$$= \left(\frac{1}{3} + \frac{7}{2}\right)$$

$$= \frac{1}{3} + \frac{21}{6}$$

$$= \frac{1}{3} + \frac{7}{2}$$

$$= \frac{1}{3} + \frac{21}{6}$$

$$= \frac{1}{3} + \frac{7}{2}$$

$$= \frac{1 + 21}{3 + 6}$$

$$= \frac{22}{9}$$

Therefore, the area under the curve $f(x) = x^2 - 3x + 5$ in the interval $[0, 1]$ is $\frac{22}{9}$ square units.