_ , - ,1 ^ ,0 dx x^2 + 2x + 2 = - Vidyadip

MCQ

$\int_{\, - \,1}^{\,0} {\frac{{dx}}{{{x^2} + 2x + 2}} = } $

A
$0$
✓
$\pi /4$
C
$\pi /2$
D
$ - \pi /4$

✓

Answer

Correct option: B.

$\pi /4$

b
(b) $I = \int_{ - 1}^0 {\frac{{dx}}{{{x^2} + 2x + 2}}} $$ = \int_{ - 1}^0 {\frac{{dx}}{{{{(x + 1)}^2} + 1}}} $

$ = [{\tan ^{ - 1}}(x + 1)]_{ - 1}^0$

$ = [{\tan ^{ - 1}}1 - {\tan ^{ - 1}}0] = \frac{\pi }{4}$.

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 7.2 definite integral→7.2 definite integral — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

$\int_{}^{} {\sqrt {1 - \sin 2x} \;} dx = ........,\;\;x \in (0,\;\pi /4)$

A rea bounded by the circle x² + y² = 1 and the curve | x | + | y | = 1 is:

$2\pi$
$\pi-2$
$\pi$
$\pi+3$

Let $\text{A}=\{\text{x}\in\text{R}:\text{x}\geq1\}.$ The inverse of the function f : A → A given by $\text{f(x)}=2^{\text{x}(\text{x}-1)},$ is:

$\big(\frac{1}{2}\big)^{\text{x}(\text{x}-1)}$
$\frac{1}{2}\big\{1+\sqrt{1+4\log_2\text{x}}\big\}$
$\frac{1}{2}\big\{1-\sqrt{1+4\log_2\text{x}}\big\}$
$\text{Not defined}$

The records of a hospital show that $10\%$ of the cases of a certain disease are fatal. If $6$ patients are suffering from the disease, then the probability that only three will die is

Let $f ( x )=\int \frac{\sqrt{ x }}{(1+ x )^{2}} d x ( x \geq 0) .$ Then $f (3)- f (1)$ is equal to

The constraints $x+y \leq 4,3 x+3 y \geq 18, x \geq 0, y \geq 0$ defines on

If $\frac{{dy}}{{dx}} = {e^{ - 2y}}$ and $y = 0$ when $x = 5,$ the value of $x$ for $y = 3$ is

What is the maximum value of function $f(x)=\sin x+\cos x$ ?

Range of $f(x) = sin^{-1} (\sqrt {x^2 + x +1})$ is -

$\int_{}^{} {{e^{ - 2x}}\sin 3x\;dx = } $