સંકલન int limits_1^3 left((x-2)^4 sin^3(x-2) | vedclass

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Question

(B) ધારો કે $I = \int \limits_1^3 \left((x-2)^4 \sin^3(x-2) + (x-2)^{2019} + 1\right) dx$ $(i)$.ગુણધર્મ $\int \limits_a^b f(x) dx = \int \limits_a^b f

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) ધારો કે $I = \int \limits_1^3 \left((x-2)^4 \sin^3(x-2) + (x-2)^{2019} + 1\right) dx$ $(i)$.ગુણધર્મ $\int \limits_a^b f(x) dx = \int \limits_a^b f(a+b-x) dx$ નો ઉપયોગ કરતા,જ્યાં $a=1$ અને $b=3$,આપણને $a+b-x = 4-x$ મળે છે.$x$ ને $4-x$ વડે બદલતા:$I = \int \limits_1^3 \left((4-x-2)^4 \sin^3(4-x-2) + (4-x-2)^{2019} + 1\right) dx$$I = \int \limits_1^3 \left((2-x)^4 \sin^3(2-x) + (2-x)^{2019} + 1\right) dx$કારણ કે $(2-x)^4 = (x-2)^4$ અને $\sin^3(2-x) = -\sin^3(x-2)$,તથા $(2-x)^{2019} = -(x-2)^{2019}$,તેથી:$I = \int \limits_1^3 \left(-(x-2)^4 \sin^3(x-2) - (x-2)^{2019} + 1\right) dx$ $(ii)$.$(i)$ અને $(ii)$ નો સરવાળો કરતા:$2I = \int \limits_1^3 \left(((x-2)^4 \sin^3(x-2) + (x-2)^{2019} + 1) + (-(x-2)^4 \sin^3(x-2) - (x-2)^{2019} + 1)\right) dx$$2I = \int \limits_1^3 (1 + 1) dx = \int \limits_1^3 2 dx = 2[x]_1^3 = 2(3-1) = 4$.તેથી,$I = 2$.

સંકલન $\int \limits_1^3 \left((x-2)^4 \sin^3(x-2) + (x-2)^{2019} + 1\right) dx$ નું મૂલ્ય શોધો.

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