The value of the definite integral int_{2}^{4 | vedclass

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Question

(B) Let $I = \int_{2}^{4} (x(3 - x)(4 + x)(6 - x)(10 - x) + \sin x) dx$  $(1)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$

Vedclass · Accepted Answer

$\cos 2 - \cos 4$

Vedclass · Answer

(B) Let $I = \int_{2}^{4} (x(3 - x)(4 + x)(6 - x)(10 - x) + \sin x) dx$  $(1)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$,where $a = 2$ and $b = 4$,we have $a + b - x = 6 - x$.Substituting $x$ with $6 - x$ in the integral:$I = \int_{2}^{4} ((6 - x)(3 - (6 - x))(4 + (6 - x))(6 - (6 - x))(10 - (6 - x)) + \sin(6 - x)) dx$$I = \int_{2}^{4} ((6 - x)(x - 3)(10 - x)x(4 + x) + \sin(6 - x)) dx$Note that the polynomial part $(6 - x)(x - 3)(10 - x)x(4 + x)$ is the negative of the original polynomial $x(3 - x)(4 + x)(6 - x)(10 - x)$.Wait,let us re-evaluate the polynomial: $P(x) = x(3 - x)(4 + x)(6 - x)(10 - x)$.$P(6 - x) = (6 - x)(3 - (6 - x))(4 + (6 - x))(6 - (6 - x))(10 - (6 - x)) = (6 - x)(x - 3)(10 - x)x(4 + x) = -P(x)$.Thus,adding the two forms of $I$:$2I = \int_{2}^{4} (P(x) + \sin x + P(6 - x) + \sin(6 - x)) dx = \int_{2}^{4} (\sin x + \sin(6 - x)) dx$$2I = [-\cos x - \cos(6 - x)]_{2}^{4}$$2I = (-\cos 4 - \cos 2) - (-\cos 2 - \cos 4) = 0$.Re-checking the integral: The polynomial term is indeed odd about the midpoint $x=3$. The integral of the polynomial part is $0$. $I = \int_{2}^{4} \sin x dx = [-\cos x]_{2}^{4} = \cos 2 - \cos 4$.

The value of the definite integral $\int_{2}^{4} (x(3 - x)(4 + x)(6 - x)(10 - x) + \sin x) dx$ equals

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