Question

-1436.08379036296
the answer above is not correct.
use the fundamental theorem of calculus to evaluate (if it exists)
$int_{-pi}^{pi} f(x) dx$,
where
$f(x)=\

$$\begin{cases}10x & \\text{if } -\\pi \\leq x \\leq 0 \\\\ 1\\sin(x) & \\text{if } 0 < x \\leq \\pi\\end{cases}$$

$
if the integral does not exist, type \dne\ as your answer.
$-\frac{3\pi^{6}}{2}+6$
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Explanation:

Step1: Split integral at x=0

$$\int_{-\pi}^{\pi} f(x) dx = \int_{-\pi}^{0} 10x dx + \int_{0}^{\pi} \sin(x) dx$$

Step2: Evaluate first integral

Antiderivative of $10x$ is $5x^2$.
$$\int_{-\pi}^{0} 10x dx = 5x^2 \bigg|_{-\pi}^{0} = 5(0)^2 - 5(-\pi)^2 = 0 - 5\pi^2 = -5\pi^2$$

Step3: Evaluate second integral

Antiderivative of $\sin(x)$ is $-\cos(x)$.
$$\int_{0}^{\pi} \sin(x) dx = -\cos(x) \bigg|_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) + 1 = 1 + 1 = 2$$

Step4: Sum the two results

$$-5\pi^2 + 2$$

Answer:

$-5\pi^2 + 2$