int_0^2 x^3(2-x)^4 , dx = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) माना $I = \int_0^2 x^3(2-x)^4 \, dx$. गुणधर्म $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$ का उपयोग करने पर: $I = \int_0^2 (2-x)^3(2-(2-x))^4 \,

Vedclass · Accepted Answer

$\frac{32}{35}$

Vedclass · Answer

(D) माना $I = \int_0^2 x^3(2-x)^4 \, dx$. गुणधर्म $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$ का उपयोग करने पर: $I = \int_0^2 (2-x)^3(2-(2-x))^4 \, dx = \int_0^2 (2-x)^3 x^4 \, dx$. पदों का विस्तार करने पर: $I = \int_0^2 (8 - 12x + 6x^2 - x^3) x^4 \, dx = \int_0^2 (8x^4 - 12x^5 + 6x^6 - x^7) \, dx$. प्रत्येक पद का समाकलन करने पर: $I = \left[ \frac{8x^5}{5} - \frac{12x^6}{6} + \frac{6x^7}{7} - \frac{x^8}{8} \right]_0^2$. $I = \left[ \frac{8(32)}{5} - 2(64) + \frac{6(128)}{7} - \frac{256}{8} \right]$. $I = \frac{256}{5} - 128 + \frac{768}{7} - 32 = \frac{256}{5} + \frac{768}{7} - 160$. $I = \frac{1792 + 3840 - 5600}{35} = \frac{32}{35}$.

$\int_0^2 x^3(2-x)^4 \, dx = $

Similar Questions

$\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 + \cos x} = \dots$

$\int\limits_0^\pi {\frac{{\sin \left( {n + \frac{1}{2}} \right)x}}{{\sin \frac{x}{2}}}} \,dx$,$(n \in N)$ का मान ज्ञात कीजिए।

$\int_0^\pi x \log(\sin x) \, dx = $

यदि $I = \int\limits_0^{\frac{\pi}{2}} \ln(\sin x) dx$ है,तो $\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\sin x + \cos x) dx =$

यदि $\int_{-a}^a f(x) dx = \int_0^a f(x) dx + \int_0^a g(x) dx$ है,तो $g(x) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module