If-the-function-of-f-is-continous-in-R-and-0-x-f-t-dt-x-1-t-2-f-t-dt-2x-2-4x-c-x-R-The-value-of-constant-c-is-

Question Number 83819 by jagoll last updated on 06/Mar/20

If the function of f is continous

in R and \int \underset{0}{\overset{x}{}} f (t) dt = \int \underset{x}{\overset{1}{}} t^{2} f (t) dt +

{2 x}^{2} + 4 x + c, \forall x \in R .

The value of constant c is

Answered by john santu last updated on 06/Mar/20

\frac{d}{dx} [\int \underset{0}{\overset{x}{}} f (t) dt] = \frac{d}{dx} [\int \underset{x}{\overset{1}{}} t^{2} f (t) dt +

{2 x}^{2} + 4 x + c]

f (x) = - x^{2} f (x) + 4 x + 4

(1 + x^{2}) f (x) = 4 x + 4

f (x) = \frac{4 x + 4}{x^{2} + 1} \Rightarrow f (t) = \frac{4 t + 4}{t^{2} + 1}

(1) \int \underset{0}{\overset{x}{}} \frac{2 (2 t)}{t^{2} + 1} dt + \int \underset{0}{\overset{x}{}} \frac{4}{t^{2} + 1} dt =

2 \ln ∣ x^{2} + 1 ∣ + 4 \tan^{- 1} (x)

(2) \int \underset{x}{\overset{1}{}} \frac{{4 t}^{3} + {4 t}^{2}}{t^{2} + 1} dt + {2 x}^{2} + 4 x + c =

- {2 x}^{2} - 4 x + 6 + \int \underset{x}{\overset{1}{}} \frac{4 - 4 t}{t^{2} + 1} dt + {2 x}^{2} - 4 x + c

= 6 + c + \int \underset{1}{\overset{x}{}} \frac{4}{t^{2} + 1} dt - 2 \int \underset{1}{\overset{x}{}} \frac{2 t}{t^{2} + 1} dt

= 6 + c + 4 \tan^{- 1} (x) - π - 2 \int \underset{1}{\overset{x}{}} \frac{d (t^{2} + 1)}{t^{2} + 1}

= 6 + c + {4 \tan}^{- 1} (x) - π - 2 \ln ∣ x^{2} + 1 ∣ - \ln (4)

∴ c = 6 - π - \ln 4

Commented by jagoll last updated on 06/Mar/20

waw \dots thank you very much

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