The value of the integral int limits_1^3 left | vedclass

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(B) Let $I = \int \limits_1^3 \left((x-2)^4 \sin^3(x-2) + (x-2)^{2019} + 1\right) dx$ $(i)$.Using the property $\int \limits_a^b f(x) dx = \int \limit

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$2$

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(B) Let $I = \int \limits_1^3 \left((x-2)^4 \sin^3(x-2) + (x-2)^{2019} + 1\right) dx$ $(i)$.Using the property $\int \limits_a^b f(x) dx = \int \limits_a^b f(a+b-x) dx$,where $a=1$ and $b=3$,we have $a+b-x = 4-x$.Substituting $x$ with $4-x$:$I = \int \limits_1^3 \left((4-x-2)^4 \sin^3(4-x-2) + (4-x-2)^{2019} + 1\right) dx$$I = \int \limits_1^3 \left((2-x)^4 \sin^3(2-x) + (2-x)^{2019} + 1\right) dx$Since $(2-x)^4 = (x-2)^4$ and $\sin^3(2-x) = -\sin^3(x-2)$,and $(2-x)^{2019} = -(x-2)^{2019}$,we get:$I = \int \limits_1^3 \left(-(x-2)^4 \sin^3(x-2) - (x-2)^{2019} + 1\right) dx$ $(ii)$.Adding $(i)$ and $(ii)$:$2I = \int \limits_1^3 \left(((x-2)^4 \sin^3(x-2) + (x-2)^{2019} + 1) + (-(x-2)^4 \sin^3(x-2) - (x-2)^{2019} + 1)\right) dx$$2I = \int \limits_1^3 (1 + 1) dx = \int \limits_1^3 2 dx = 2[x]_1^3 = 2(3-1) = 4$.Therefore,$I = 2$.

The value of the integral $\int \limits_1^3 \left((x-2)^4 \sin^3(x-2) + (x-2)^{2019} + 1\right) dx$ is

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