Mathematics

# The value of $\int {\left( {x - 1} \right){e^{ - x}}}$ is equal to

$- x{e^{ - x}} + C$

##### SOLUTION
$\int (x-1) e^{-x} \ dx$.
$-x = t - dx= dt$
$= \int (t+1) e^{t} \ dt$
$= (t+1 )e^{t} - \int e^{t} \ dt$ [Integration by  parts].
$= (t+1) e^{t} - e^{t} +c$
$= te^{t} + c$
$= - xe^{-x}+ c$.
$C$ is correct.

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle\lim_{n\rightarrow \infty}\dfrac{3}{n}\left\{1+\sqrt{\dfrac{n}{n+3}}+\sqrt{\dfrac{n}{n+6}}+\sqrt{\dfrac{n}{n+9}}+...….+\sqrt{\dfrac{n}{n+3(n-1)}}\right\}=?$
• A. Does not exist
• B. $1$
• C. $3$
• D. $2$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
The value of $\displaystyle \lim_{n\rightarrow \infty } \sum_{r=1}^{4n} \dfrac{\sqrt{n}}{\sqrt{r}(3\sqrt{r}+4\sqrt{n})^{2}}$ is equal to
• A. $\displaystyle\dfrac{1}{35}$
• B. $\displaystyle \dfrac{1}{14}$
• C. $\displaystyle \dfrac{1}{5}$
• D. $\displaystyle \dfrac{1}{10}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\int_0^1 {\dfrac{{dx}}{{x + \sqrt x }}}$ equals
• A. $ln\ 2$
• B. $2\ln 2-2$
• C. $2ln\ 2-1$
• D. $2ln\ 2$

1 Verified Answer | Published on 17th 09, 2020

Q4 Passage Hard
Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
Case I If $m>0$, then put $\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$
Case II If $p>0$, then put $\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$
Case III If quadratic equation $mx^2+nx+p=0$ has real roots $\alpha$ and $\beta$, then put $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
What is $\displaystyle \int \dfrac{dx}{2x^2 - 2x + 1}$ equal to ?
• A. $\dfrac{\tan^{-1} (2x - 1)}{2} + c$
• B. $2 \tan^{-1} (2x - 1) += c$
• C. $\dfrac{\tan^{-1} (2x + 1)}{2} + c$
• D. $\tan^{-1} (2x - 1) + c$