समाकल int_{1}^{5}[|x-3|+|1-x|] dx का मान किसक | vedclass

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(C) माना $I = \int_{1}^{5} [|x-3| + |1-x|] dx$. चूँकि $x \in [1, 5]$, इसलिए $|1-x| = x-1$ होगा। अतः, $I = \int_{1}^{5} |x-3| dx + \int_{1}^{5} (x-1) d

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) माना $I = \int_{1}^{5} [|x-3| + |1-x|] dx$. चूँकि $x \in [1, 5]$, इसलिए $|1-x| = x-1$ होगा। अतः, $I = \int_{1}^{5} |x-3| dx + \int_{1}^{5} (x-1) dx$. प्रथम भाग के लिए, $\int_{1}^{5} |x-3| dx = \int_{1}^{3} -(x-3) dx + \int_{3}^{5} (x-3) dx$. $= \int_{1}^{3} (3-x) dx + \int_{3}^{5} (x-3) dx = [3x - \frac{x^2}{2}]_{1}^{3} + [\frac{x^2}{2} - 3x]_{3}^{5}$. $= (9 - 4.5) - (3 - 0.5) + (12.5 - 15) - (4.5 - 9) = 4.5 - 2.5 - 2.5 + 4.5 = 4$. दूसरे भाग के लिए, $\int_{1}^{5} (x-1) dx = [\frac{x^2}{2} - x]_{1}^{5} = (12.5 - 5) - (0.5 - 1) = 7.5 - (-0.5) = 8$. अतः, $I = 4 + 8 = 12$.

$x$	$1$	$2$	$3$	$4$
$y$	$0.7111$	$0.7222$	$0.7333$	$0.7444$

समाकल $\int_{1}^{5}[|x-3|+|1-x|] dx$ का मान किसके बराबर है?

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