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, so the set is $\{2\}$.

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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