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#