JEE Mains · Maths · STD 12 - 7.2 definite integral
Let \(f: R \rightarrow R\) be defined as \(f(x)=e^{-x} \sin x\). If \(F :[0,1] \rightarrow R\) is a differentiable function such that \(F ( x )=\int_{0}^{ x } f ( t ) dt ,\) then the value of \(\int_{0}^{1}\left( F ^{\prime}( x )+ f ( x )\right) e ^{ x } dx\) lies in the interval
- A \(\left[\frac{327}{360}, \frac{329}{360}\right]\)
- B \(\left[\frac{330}{360}, \frac{331}{360}\right]\)
- C \(\left[\frac{331}{360}, \frac{334}{360}\right]\)
- D \(\left[\frac{335}{360}, \frac{336}{360}\right]\)
Answer & Solution
Correct Answer
(B) \(\left[\frac{330}{360}, \frac{331}{360}\right]\)
Step-by-step Solution
Detailed explanation
\(f ( x )= e ^{- x } \sin x\) Now, \(F ( x )=\int_{0}^{ x } f ( t ) d t \quad \Rightarrow F ^{\prime}( x )= f ( x )\) \(I =\int_{0}^{1}\left( F ^{\prime}( x )+ f ( x )\right) e ^{ x } dx =\int_{0}^{1}( f ( x )+ f ( x )) \cdot e ^{ x } dx\)…
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