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Exercises · Q12

Q.Evaluate ∫14(2x+1) dx\displaystyle\int_{1}^{4} (2x+1)\,dx directly, and again by splitting the interval at x=2x=2 using the additivity property. Confirm both methods agree.

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Antiderivative: F(x)=x2+xF(x)=x^{2}+x.

Method 1 — direct evaluation:

F(4)=16+4=20,F(1)=1+1=2F(4)=16+4=20, \qquad F(1)=1+1=2

∫14(2x+1) dx=F(4)−F(1)=20−2=18\int_{1}^{4}(2x+1)\,dx = F(4)-F(1) = 20-2 = 18

Method 2 — split at c=2c=2 using additivity:

∫12(2x+1) dx=F(2)−F(1)=(4+2)−2=6−2=4\int_{1}^{2}(2x+1)\,dx = F(2)-F(1) = (4+2)-2 = 6-2 = 4

∫24(2x+1) dx=F(4)−F(2)=20−6=14\int_{2}^{4}(2x+1)\,dx = F(4)-F(2) = 20-6 = 14

∫12(2x+1) dx+∫24(2x+1) dx=4+14=18\int_{1}^{2}(2x+1)\,dx + \int_{2}^{4}(2x+1)\,dx = 4+14 = 18 …

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