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MathematicsDefinite IntegrationDerivatives (Newton- Leibnitz)Medium2 minPYQ_2024
MathematicsMediumnumerical

LetS=1, andf:Sbe defined asfx=1xet1112t15t27t3122t1061dt. Letp=Sum of square of the values ofx, wherefxattains local maxima onS. andq=Sum of the values ofx, wherefxattains local minima onS. Then, the value ofp2+2qis ________

Question diagram: Let S = − 1 , ∞ and f : S → ℝ be defined as f x = ∫ − 1 x e
Answer:
27.00
Solution:

Given,

fx=1xet1112t15t27t3122t1061dt

Now, differentiating the above integral using Newton Leibnitz's theorem, we get,

f'x=ex1112x15x27x3122x1061

Now, from the above diagram using first derivative test we get,

Local minima at x=12, x=5 

Local maxima at x=0, x=2

So, p=0+4=4, q=12+5=112

Hence, p2+2q=16+11=27

Stream:JEESubject:MathematicsTopic:Definite IntegrationSubtopic:Derivatives (Newton- Leibnitz)
2mℹ️ Source: PYQ_2024

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