GATE 2024 ES – Question 49
Find area under $y=x^2+x$ above the x-axis between x=1 and x=2,to two decimals.
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Correct answer: 3.70 to 3.90
Explanation
The curve is positive throughout[1,2],so area is its definite integral: $$A=\int_1^2(x^2+x)dx=[x^3/3+x^2/2]_1^2=(8/3+2)-(1/3+1/2)=23/6.$$ Therefore **3.83 square units**. No absolute-value splitting is needed on this positive interval.