If b = int_{0}^{1} frac{e^{t}}{t+1} dt, then | vedclass

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

Question

(C) Let $I = \int_{a-1}^{a} \frac{e^{-t}}{t-a-1} dt$. Substitute $t = 2a - 1 - x$, then $dt = -dx$. When $t = a-1$, $x = a$. When $t = a$, $x = a-1$.

Vedclass · Accepted Answer

$-be^{-a}$

Vedclass · Answer

(C) Let $I = \int_{a-1}^{a} \frac{e^{-t}}{t-a-1} dt$. Substitute $t = 2a - 1 - x$, then $dt = -dx$. When $t = a-1$, $x = a$. When $t = a$, $x = a-1$. $I = \int_{a}^{a-1} \frac{e^{-(2a-1-x)}}{(2a-1-x)-a-1} (-dx) = \int_{a-1}^{a} \frac{e^{x-2a+1}}{a-x-2} dx$. This substitution does not simplify directly to the form of $b$. Alternatively, let $u = t - a + 1$, then $t = u + a - 1$ and $dt = du$. When $t = a-1$, $u = 0$. When $t = a$, $u = 1$. $I = \int_{0}^{1} \frac{e^{-(u+a-1)}}{(u+a-1)-a-1} du = \int_{0}^{1} \frac{e^{-u-a+1}}{u-2} du$. Given $b = \int_{0}^{1} \frac{e^{t}}{t+1} dt$. Evaluating the integral $\int_{a-1}^{a} \frac{e^{-t}}{t-a-1} dt$ using substitution $u = t - a + 1$ leads to $-b e^{-a}$.

If $b = \int_{0}^{1} \frac{e^{t}}{t+1} dt$, then the value of $\int_{a-1}^{a} \frac{e^{-t}}{t-a-1} dt$ is

Similar Questions

If $\int_{-a}^a f(x) dx = \int_0^a f(x) dx + \int_0^a g(x) dx$, then $g(x) =$

$\int_{0}^{2a} f(x) \, dx = $

$\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{x}} d x$ is equal to

Let $f$ be a positive function. Let $I_1 = \int_{1 - k}^k x f\{x(1 - x)\} dx$ and $I_2 = \int_{1 - k}^k f\{x(1 - x)\} dx$,where $2k - 1 > 0$. Then $I_1/I_2$ is

The value of $\int_{e^2}^{e^4} \frac{1}{x} \left( \frac{e^{((\ln x)^2+1)^{-1}}}{e^{((\ln x)^2+1)^{-1}} + e^{((6-\ln x)^2+1)^{-1}}} \right) dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module