MHT CET · Maths · Definite Integration
\(\int_{0}^{1}\left(1-\frac{x}{1 !}+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\cdots\right.\) upto \(\left.\infty\right) e^{2 x} d x=\)
- A \(e^{2}\)
- B \(e-1\)
- C \(e+1\)
- D \(e\)
Answer & Solution
Correct Answer
(B) \(e-1\)
Step-by-step Solution
Detailed explanation
(B)
\(\int_{0}^{1}\left(1-\frac{x}{1 !}+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\ldots \infty\right) e^{2 x} d x\)
\(=\int_{0}^{1} e^{-x} e^{2 x} d x=\int_{0}^{1} e^{x} d x\)
\(=\left[e^{x}\right]_{0}^{1}=e^{1}-e^{0}=e-1\)
\(\int_{0}^{1}\left(1-\frac{x}{1 !}+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\ldots \infty\right) e^{2 x} d x\)
\(=\int_{0}^{1} e^{-x} e^{2 x} d x=\int_{0}^{1} e^{x} d x\)
\(=\left[e^{x}\right]_{0}^{1}=e^{1}-e^{0}=e-1\)
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