Step 1: ∫152xdx
∫2xdx=x2+C
[x2]15=52−12
52=25,12=1
25−1=24
∫152xdx=24
24
Step 2: ∫15(2x+3)dx
∫(2x+3)dx=x2+3x+C
[x2+3x]15=(52+3⋅5)−(12+3⋅1)
52+3⋅5=25+15=40
12+3⋅1=1+3=4
40−4=36
∫15(2x+3)dx=36
36
Step 3: f(x)=x2−3x+2, ∫15f(x)dx=∫15(x2−3x+2)dx
∫(x2−3x+2)dx=31x3−23x2+2x+C
[31x3−23x2+2x]15
At x=5:
31(125)−23(25)+2(5)=3125−275+10
Common denominator 6:
3125=6250,275=6225,10=660
6250−6225+660=6250−225+60=685
At x=1:
31(1)−23(1)+2(1)=31−23+2
Common denominator 6:
31=62,23=69,2=612
62−69+612=62−9+12=65
685−65=680=340
∫15(x2−3x+2)dx=340
\dfrac{40{3}}