Integrals

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New answer posted•a year ago

Integrals Class 12th

The value of the integral $\int_{1}^{3} [x^{2} - 2 x - 2] d x,$  where [x] denotes the greatest integer less than or equal to x, is

9 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$l = \int_{1}^{3} [x^{2} - 2 x + 1 - 3] d x = \int_{1}^{3} [{(x - 1)}^{2} - 3] d x = \int_{1}^{3} [{(x - 1)}^{2}] d x - 3 \int_{1}^{3} d x$ .(A)

$l_{1} = \int_{1}^{3} [{(x - 1)}^{2}] d x P u t {(x - 1)}^{2} = t$

$l_{1} = \frac{1}{2} [0 \int_{0}^{1} \frac{d t}{\sqrt[]{t}} \int_{1}^{2} \frac{d t}{\sqrt[]{t}} + 2 \int_{2}^{3} \frac{d t}{\sqrt[]{t}} + 3 \int_{3}^{4} \frac{d t}{\sqrt[]{t}}] = \frac{1}{2} {{| \frac{t^{\frac{- 1}{2} + 1}}{- \frac{1}{2} + 1} |}_{1}^{2} {| + 2 \frac{t^{\frac{- 1}{2} + 1}}{- \frac{1}{2} + 1} |}_{2}^{3} {+ 3 \frac{t^{\frac{- 1}{2} + 1}}{- \frac{1}{2} + 1}]}_{3}^{4}}$ = $5 - \sqrt[]{2} - \sqrt[]{3}$

Hence from (A)

= $5 - \sqrt[]{2} - \sqrt[]{3} - \sqrt[]{3} - 6 = - 1 - \sqrt[]{2} - \sqrt[]{3}$

2^nd method

\int_{1}^{3} [{(x - 1)}^{2}] d x - 6 . . . . . . . . . . . . . . (A)

From (A), $l = 5 - \sqrt[]{2} - \sqrt[]{3} - 6 = - 1 - \sqrt[]{2} - \sqrt[]{3}$

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New answer posted•a year ago

Integrals Class 12th

The value of the integral $\int \frac{s i n θ s i n θ (s i n^{6} θ + s i n^{4} θ + s i n^{2} θ) \sqrt[]{2 s i n^{4} θ + 3 s i n^{2} θ + 6}}{1 - c o s 2 θ} d θ$

(where c is a constant of integration)

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Let sinθ= t $\int \frac{s i n θ (2 s i n θ . c o s θ) (s i n^{6} θ + s i n^{4} θ + s i n^{2} θ) \sqrt[]{2 s i n^{4} θ + 3 s i n^{2} θ + 6}}{2 s i n^{2} θ}$ dθ

sinθ = t

cos θ. dθ = dt

$\int \frac{u^{1 / 2}}{12} d u = \frac{u^{3 / 2}}{18} + C = \frac{{(2 t^{6} + 3 t^{4} + 6 t^{2})}^{3 / 2}}{18} + C = \frac{{(2 s i n^{6} θ + 3 s i n^{4} θ + 6 s i n^{2} θ)}^{3 / 2}}{18} + C$

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New answer posted•a year ago

Integrals Class 12th

$\underset{n \to \infty}{l i m} (1 + \frac{1 + \frac{1}{2} + . . . . . . . . . . + \frac{1}{n}}{n^{2}}) i s e q u a l t o :$

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$\underset{n \to \infty}{l i m} {(1 + \frac{1 + \frac{1}{2} + . . . . . . . + \frac{1}{n}}{n^{2}})}^{n}$ limit is in the form of $1^{\infty}$

$l = e x p (\underset{n \to \infty}{l i m} \frac{1 + \frac{1}{2} + \frac{1}{3} + . . . . . + \frac{1}{n}}{n^{2}})$

$0 \leq 1 + \frac{1}{2} + \frac{1}{3} + . . . . . + \frac{1}{n} \leq 1 + \frac{1}{\sqrt[]{2}} + \frac{1}{\sqrt[]{3}} + . . . . + \frac{1}{\sqrt[]{n}} \leq 2 \sqrt[]{n} - 1$

Taking limit $(n \to \infty)$

$l$ = exp (0) (from sandwich)

$l = 1$

Second Method :

1 + \frac{1}{2} + \frac{1}{3} + . . . . + \frac{1}{n} \geq l n (n + 1) . . . . . . . (i)

1 + \frac{1}{2} + \frac{1}{3} + . . . . . + \frac{1}{n} \leq 1 + \int_{1}^{2} \frac{1}{2} d x \leq 1 + l n n . . . . . . . . . . (i i)

From (i) & (ii)

$l n (n + 1) \leq 1 + \frac{1}{2} + \frac{1}{3} + . . . . + \frac{1}{n} \leq 1 + I n n, \forall n \in N, n \geq 2$ Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$l = \int \frac{8 x^{3} + 20 x^{2}}{x^{4} + 5 x^{3} - 7 x^{2}} d x = 4 \int \frac{2 x + 5}{x^{2} + 5 x - 7} d x = 4 l n | x^{2} + 5 x - 7 | + c$

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New answer posted•a year ago

Integrals Class 12th

If In = $\int_{\frac{π}{4}}^{\frac{π}{2}} c o t^{n} x d x, t h e n :$

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$l_{n} = \int_{π / 4}^{π / 2} c o t^{n} x d x = \int_{π / 4}^{π / 2} c o t^{n - 2} x (c o s e c^{2} x - 1) d x$

$= {[\frac{- c o t^{n - 1} x}{n - 1}]}_{π / 4}^{π / 2} - l_{n - 2} = \frac{1}{n - 1} - l_{n - 2}$

$l_{n} + l_{n - 2} = \frac{1}{n - 1}$

$\begin{array}{l} n = 4 \Rightarrow l_{4} + l_{2} = \frac{1}{3} \\ n = 5 \Rightarrow l_{5} + l_{3} = \frac{1}{4} \\ n = 6 \Rightarrow l_{6} + l_{4} = \frac{1}{5} \end{array}} \Rightarrow \frac{1}{l_{2} + l_{4}}, \frac{1}{l_{3} + l_{5}}, \frac{1}{l_{4} + l_{6}}$ are in A.P.

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New answer posted•a year ago

Integrals Class 12th

The area (in sq. units) of the region bounded by the curves $x^{2} + 2 y - 1 = 0, y^{2} + 4 x - 4 = 0$ and y² – 4x – 4 = 0, in the upper half plane is……….

9 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$x^{2} = - 2 (y - \frac{1}{2})$

y² = -4 (x – 1)

Required area = $2 \int_{- 1}^{0} [2 \sqrt[]{x + 1} - \frac{1 - x^{2}}{2}] d x$

$= 2 {[\frac{4}{3} {(x + 1)}^{\frac{3}{2}} - \frac{1}{2} (x - \frac{x^{3}}{3})]}_{- 1}^{0}$

= 2

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New answer posted•a year ago

Integrals Class 12th

If $\int_{0}^{100 π} \frac{s i n^{2} x}{(\frac{x}{π} - [\frac{x}{π}])} d x = \frac{α π^{3}}{1 + 4 π^{2}}, α \in R,$  where [x] is the greatest integer less than or equal to x, the value of a is:

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$\int_{}^{} \frac{s i n^{2} x d x}{e^{{\frac{x}{π}}}} = l$

$f (x) = \frac{s i n^{2} x}{e^{{\frac{x}{π}}}}$ is periodic with period =

$l = 100 \int_{0}^{π} \frac{s i n^{2} x}{e^{x / π}} d x .$

$= 50 \int_{0}^{π} \frac{1 - c o s 2 x}{e^{x / π}} d x = 50 \int_{0}^{π} e^{- x / π} d x - 50 \int_{0}^{π} e^{- x / π} c o s 2 x d x$

$= 50 π (1 - \frac{1}{e}) [1 - \frac{1}{1 + 4 π^{2}}] = \frac{50 (1 - \frac{1}{e}) 4 π^{3}}{1 + 4 π^{2}}$

$∴ α = 200 (1 - \frac{1}{e})$

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New answer posted•a year ago

Integrals Class 12th

For $k \in N,$  let $\frac{1}{α (α + 1) (α + 2) . . . . . . (α + 20)} = \sum_{k = 0}^{20} \frac{A_{k}}{a + k},$  where a > 0, Then the value of $100 {(\frac{A_{14} + A_{15}}{A_{13}})}^{2}$  is equal to…………….

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\frac{1}{α (α + 1) (α + 2) . . . . . . . . . . (α + 20)} = \frac{A_{0}}{α} + \frac{A_{1}}{α + 1} + \frac{A_{2}}{α + 2} + . . . . . . + \frac{A_{20}}{α + 20}$

Solving by partial fraction, we get

$A_{13} = \frac{- 1}{14! 7!}, A_{14} = \frac{- 1}{14! 6!} a n d A_{15} = \frac{- 1}{14! 5!}$

$∴ \frac{A_{14} + A_{15}}{A_{13}} = \frac{3}{10} \Rightarrow {(\frac{A_{14} + A_{15}}{A_{13}})}^{2} = {(\frac{3}{10})}^{2} \Rightarrow 100 {(\frac{A_{14} + A_{15}}{A_{13}})}^{2} = 9$