यदि int f(x) dx = varphi(x) है,तो int x^5 f(x | vedclass

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(C) दिया गया है $\int f(x) dx = \varphi(x)$.माना $I = \int x^5 f(x^3) dx$.$t = x^3$ प्रतिस्थापित करने पर,$dt = 3x^2 dx$,जिसका अर्थ है $x^2 dx = \frac{

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$\frac{1}{3} x^3 \varphi(x^3) - \int x^2 \varphi(x^3) dx + c$

Vedclass · Answer

(C) दिया गया है $\int f(x) dx = \varphi(x)$.माना $I = \int x^5 f(x^3) dx$.$t = x^3$ प्रतिस्थापित करने पर,$dt = 3x^2 dx$,जिसका अर्थ है $x^2 dx = \frac{1}{3} dt$.साथ ही,$x^5 dx = x^3 \cdot x^2 dx = t \cdot \frac{1}{3} dt$.अतः,$I = \int t f(t) \cdot \frac{1}{3} dt = \frac{1}{3} \int t f(t) dt$.खंडशः समाकलन (Integration by parts) का उपयोग करने पर,$\int u dv = uv - \int v du$,जहाँ $u = t$ और $dv = f(t) dt$ (अतः $du = dt$ और $v = \varphi(t)$).$I = \frac{1}{3} [t \varphi(t) - \int \varphi(t) dt] + c$.अब $t = x^3$ वापस रखने पर:$I = \frac{1}{3} [x^3 \varphi(x^3) - \int \varphi(x^3) \cdot (3x^2 dx)] + c$.$I = \frac{1}{3} x^3 \varphi(x^3) - \int x^2 \varphi(x^3) dx + c$.

यदि $\int f(x) dx = \varphi(x)$ है,तो $\int x^5 f(x^3) dx = $

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