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

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Answer 1 · 2015-12-10T22:32:02

$23/6$ 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

$int_0^1(x^2-3x+5)dx$

$=[x^3/3-3x^2/2+5x]_0^1$

$=1/3-3/2+5$

$=23/6$

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