The value of alpha for which 4 alpha int_{-1} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the integral $I = \int_{-1}^{2} e^{-\alpha |x|} dx$. Since $|x| = -x$ for $x < 0$ and $|x| = x$ for $x \ge 0$,we split the integral: $I = \i

Vedclass · Accepted Answer

$\log_{e} 2$

Vedclass · Answer

(C) Given the integral $I = \int_{-1}^{2} e^{-\alpha |x|} dx$. Since $|x| = -x$ for $x $I = \int_{-1}^{0} e^{\alpha x} dx + \int_{0}^{2} e^{-\alpha x} dx$. Evaluating the integrals: $I = \left[ \frac{e^{\alpha x}}{\alpha} \right]_{-1}^{0} + \left[ \frac{e^{-\alpha x}}{-\alpha} \right]_{0}^{2} = \left( \frac{1}{\alpha} - \frac{e^{-\alpha}}{\alpha} \right) + \left( \frac{1}{\alpha} - \frac{e^{-2\alpha}}{\alpha} \right) = \frac{2 - e^{-\alpha} - e^{-2\alpha}}{\alpha}$. Now,substitute this into the given equation $4\alpha I = 5$: $4\alpha \left( \frac{2 - e^{-\alpha} - e^{-2\alpha}}{\alpha} \right) = 5$. $4(2 - e^{-\alpha} - e^{-2\alpha}) = 5 \Rightarrow 8 - 4e^{-\alpha} - 4e^{-2\alpha} = 5$. $4e^{-2\alpha} + 4e^{-\alpha} - 3 = 0$. Let $t = e^{-\alpha}$. Then $4t^2 + 4t - 3 = 0$. Solving the quadratic equation: $t = \frac{-4 \pm \sqrt{16 - 4(4)(-3)}}{2(4)} = \frac{-4 \pm \sqrt{64}}{8} = \frac{-4 \pm 8}{8}$. Since $t = e^{-\alpha} > 0$,we take $t = \frac{4}{8} = \frac{1}{2}$. $e^{-\alpha} = \frac{1}{2} \Rightarrow e^{\alpha} = 2 \Rightarrow \alpha = \log_{e} 2$.

The value of $\alpha$ for which $4 \alpha \int_{-1}^{2} e^{-\alpha |x|} dx = 5$ is:

Similar Questions

If $f(x) = \int_{-1}^{x} |t| dt$,then for any $x \geq 0$,$f(x)$ is equal to

The value of $\int_{a}^{b} \frac{x}{|x|} dx$,where $a < b < 0$,is

Evaluate the definite integral $\int_{-2}^{2.24} [x] \, dx$,where $[x]$ denotes the greatest integer function.

$\int_1^2 \frac{x^4-1}{x^6-1} d x=$

$\int_0^3 |x^2 - 3x + 2| dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module