^e_1 dx= 1. 1 2. e - 1 3. e + 1 4. 0 - Vidyadip

Question

$\int\limits^\text{e}_1\log\text{x}\text{ dx}=$

1
e - 1
e + 1
0

✓

Answer

$1$

Solution:
$\int\limits^\text{e}_1\log\text{x}\text{ dx}$
$=\int\limits^\text{e}_1\log\text{x}\text{ x}^0\text{dx}$
$=\big[\text{x}\log\text{x}\big]^\text{e}_1-\int\limits^\text{e}_1\frac{1}{\text{x}}\text{dx}$
$=\big[\text{x}\log\text{x}\big]^\text{e}_1-\big[\text{x}\big]^\text{e}_1$
$=(\text{e}-0)-(\text{e}-1)$
$= \text{e}-\text{e}+1$
$=1$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from INTEGRALS→INTEGRALS — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

lf a line makes angles $\frac{\pi}{12},\frac{5\pi}{12}$ with oy, oz respectively where 0 = (0, 0, 0), then the angle made by that line with ox is:

The area $($in sq. units$)$ enclosed between the graph of $y = x^3$ and the lines $x = 0, y = 1, y = 8$ is$:$

If $\vec{\text{x}}$ is a vector in the direction of (2, -2, 1) of magnitude 6 and $\vec{\text{y}}$ is a vector in the direction of (1, 1, -1) of magnitude $\sqrt{3}$ then $\mid\vec{\text{x}}+2\vec{\text{y}}\mid=$

$40$
$\sqrt{35}$
$\sqrt{17}$
$2\sqrt{10}$

Area of the region bounded by rays |x| + y = 1 and X - axis is ___________.

$\frac{1}{2}$
$2$
$1$
$\frac{1}{4}$

If $f:R\rightarrow R, f(x)=2x+1 ,$ then the value of $f^{-1}(2)$ is

If $\int\limits^1_0\text{f}(\text{x})\text{dx}=1,\int\limits^1_0\text{x}\text{f}(\text{x})\text{dx}=\text{a},\int\limits^1_0\text{x}^2\text{f}(\text{x})\text{dx}=\text{a}^2,$ then $\int\limits^1_0(\text{a}-\text{x})^2\text{f(x)}\text{dx}$ equals:

$\cot ^{-1} \frac{\sqrt{1-x^2}}{x}$ equal to :

$Z = 20x_1 + 20x_2,$ subject to $\text{x}1\geq0,\text{x}_{2}\geq0,\text{x}_{1}+2\text{x}_{2}\geq8,3\text{x}_{1}+2\text{x}_{2}\geq15,5\text{x}_{1}+2\text{x}_{2}\geq20.$ The minimum value of $Z$ occurs at

Choose the correct answer from given four options in each of the Exercise:
If $\text{f(x)}==\begin{vmatrix}0&\text{x}-\text{a}&\text{x}-\text{b} \\\text{x}+\text{a} &0&\text{x}-\text{c}\\\text{x}+\text{b}&\text{x}+\text{c}&0\end{vmatrix},$ then:

f(a) = 0
f(b) = 0
f(0) = 0
f(1) = 0

The angle of intersection of the curves $xy = a^2$ and $x^2 - y^2 = 2a^{2 }$ is: