TS EAMCET · Maths · Indefinite Integration
If \(I_n=\int x^n \cdot e^{c x} d x\) for \(n \geq 1, \quad\) then \(c \cdot I_n+n \cdot I_{n-1}\) is equal to
- A \(x^n e^{c x}\)
- B \(x^n\)
- C \(e^{c x}\)
- D \(x^n+e^{c x}\)
Answer & Solution
Correct Answer
(A) \(x^n e^{c x}\)
Step-by-step Solution
Detailed explanation
Given that, \[ \begin{aligned} & I_n=\int x^n \cdot e^{c x} d x \\ & I_n=\frac{e^{c x}}{c} \cdot x^n-\int \frac{e^{c x}}{c} \cdot n x^{n-1} d x \\ & \Rightarrow \quad I_n=\frac{e^{c x} \cdot x^n}{c}-\frac{n}{c} I_{n-1} \\ & \end{aligned} \]…
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