If the primitive of f(x) = pi sin(pi x) + 2x | vedclass

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(C) Let $F(x)$ be the primitive of $f(x)$. Then $F(x) = \int f(x) \, dx = \int (\pi \sin(\pi x) + 2x - 4) \, dx$. Integrating term by term,we get $F(x

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{$2$}

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(C) Let $F(x)$ be the primitive of $f(x)$. Then $F(x) = \int f(x) \, dx = \int (\pi \sin(\pi x) + 2x - 4) \, dx$. Integrating term by term,we get $F(x) = -\cos(\pi x) + x^2 - 4x + C$. Given that $F(1) = 3$,we substitute $x = 1$: $F(1) = -\cos(\pi) + (1)^2 - 4(1) + C = 3$. Since $\cos(\pi) = -1$,we have $-(-1) + 1 - 4 + C = 3$,which simplifies to $1 + 1 - 4 + C = 3$,so $-2 + C = 3$,giving $C = 5$. Thus,$F(x) = -\cos(\pi x) + x^2 - 4x + 5$. We want to find $x$ such that $F(x) = 0$,so $-\cos(\pi x) + x^2 - 4x + 5 = 0$,or $\cos(\pi x) = x^2 - 4x + 5$. We can rewrite the right side as $(x-2)^2 + 1$. Since $\cos(\pi x) \le 1$ and $(x-2)^2 + 1 \ge 1$,the equation holds only if $\cos(\pi x) = 1$ and $(x-2)^2 + 1 = 1$. $(x-2)^2 + 1 = 1$ implies $(x-2)^2 = 0$,so $x = 2$. Checking $x = 2$ in the cosine term: $\cos(2\pi) = 1$. Thus,$x = 2$ is the only solution.

If the primitive of $f(x) = \pi \sin(\pi x) + 2x - 4$ has the value $3$ for $x = 1$,then the set of $x$ for which the primitive of $f(x)$ vanishes is:

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