\(If ƒ(x)=∫_0^xt\ sin\ t\ dt,\ then ƒ'(x)is\)
\(If ƒ(x)=∫_0^xt\ sin\ t\ dt,\ then ƒ'(x)is\)
\(cos\ x+x\ sin\ x\)
\(x\ sin\ x\)
\(x\ cos\ x\)
\(sin\ x+x\ cos\ x\)
The Correct Option is B
Solution and Explanation
\( ƒ(x)=∫_0^xt\ sin\ t\ dt\)
\(Integrating\ by\ parts, \ we \ obtain\)
\( ƒ(x)=t∫_0^x sin\ t\ dt-∫_0^x[{(\frac {d}{dt}t)∫sin\ t\ dt}]\ dt\)
= \([t(-cos\ t)]_0^x-∫_0^x(-cos\ t)dt\)
= \([-t\ cos\ t+sin\ t]_0^x\)
= \(-x\ cos\ x+sin\ x\)
\(⇒ ƒ(x)=-[{x(-sin\ x)}+cos\ x]+cos\ x\)
= \(x\ sin\ x-cos\ x+cos\ x\)
= \(x\ sin\ x\)
\(Hence, \ correct\ Answer\ is \ B.\)
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Concepts Used:
Integration by Parts
Integration by Parts is a mode of integrating 2 functions, when they multiplied with each other. For two functions ‘u’ and ‘v’, the formula is as follows:
∫u v dx = u∫v dx −∫u' (∫v dx) dx
- u is the first function u(x)
- v is the second function v(x)
- u' is the derivative of the function u(x)
The first function ‘u’ is used in the following order (ILATE):
- 'I' : Inverse Trigonometric Functions
- ‘L’ : Logarithmic Functions
- ‘A’ : Algebraic Functions
- ‘T’ : Trigonometric Functions
- ‘E’ : Exponential Functions
The rule as a diagram:
