Question
The value of the definite integral ∫10x dxx3+16 lies in the interval [a, b]. The smallest such interval is.
Answer: Option A
:
A
f(x)=xx3+16⇒f′(x)=16−2x3(x3+16)2∴f′(x)=0⇒x=2Alsof′′(2)<0⇒ The function f(x)=xx3+16 is an increasing function in [0,1], so Min f(x) = f(0) = 0 and Max f(x) = f(1) = 117
Therefore by the property m(b−a)≤∫baf(x)dx≤M(b−a)
(where m and M are the smallest and greatest values of function)
⇒0≤∫10xx3+16dx≤117
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:
A
f(x)=xx3+16⇒f′(x)=16−2x3(x3+16)2∴f′(x)=0⇒x=2Alsof′′(2)<0⇒ The function f(x)=xx3+16 is an increasing function in [0,1], so Min f(x) = f(0) = 0 and Max f(x) = f(1) = 117
Therefore by the property m(b−a)≤∫baf(x)dx≤M(b−a)
(where m and M are the smallest and greatest values of function)
⇒0≤∫10xx3+16dx≤117
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