Question

$\begin{gathered}\bf \: if \: \: \: \rm \: f :R \to \: R \: \: \: \: and \: \\ \\ \rm \: f(x) = \displaystyle \int_1^3 \: \rm \frac{1}{ |t - x| + 1} \: \: dt \\ \\ \rm \: then \: \: \displaystyle \int_1^3 \rm \: f(t) \: dt = ? \end{gathered}$​

Answer 1

Answer:

$\huge\green{\displaystyle \int_1^3f(t) \: dt = 4}$

Step-by-step explanation:

Given that

$f(x) = \displaystyle \int_1^3 \: \rm \frac{1}{ |t - x| + 1} \: \: dt$

Note that x and t are independent variables , so

$f(t) = \displaystyle \int_1^3 \: \rm \frac{1}{ |t - t| + 1} \: \: dt \\ \\ \implies \: f(t) = \displaystyle \int_1^3 \: \rm \frac{1}{ 0 + 1} \: \: dt \\ \\ \implies \: f(t) = \displaystyle \int_1^3dt = 3 - 1 \\ \\ \implies \: f(t) = 2$

Now integrate both sides from 1 to 3

$\displaystyle \int_1^3f(t)\: \: dt = \displaystyle \int_1^32 \: dt \\ = 2(3 - 1) \\ = 4$

Answer 2

$\huge \red{ \boxed{\sf \int_3^{1} f(t)dt = \pink4}}$

EXPLANATION :

$\large \sf f(x) = \int_ {3}^{1} \frac{1}{ |t - x| + 1} \: dt$

NOTE POINT :

↣ x and t are independent variables

$\large \sf f(t) = \int_ {3}^{1} \frac{1}{ |t - t| + 1 } \: dt$

$\large \sf f(t) = \int_ {3}^{1} \frac{1}{0 + 1} \: dt$

$\large \sf f(t) = \int_ {3}^{1} dt = 3 - 1 = 2$

⇒ f(t) = 2

NOW BY INTEGRATE ON BOTH SIDES BY 1 TO 3,

$\Large \sf\int_ {3}^{1} f(t) \: dt = \int_ {3}^{1} 2 \: dt$

$\sf \Large = \: 2(3 - 1)$

$2$