Evaluating Definite Integrals by Finding Antiderivatives Question - None

Question

If $f(x) = \begin{cases} x, & x < 1 \ \sin x, & x \geq 1 \end{cases}$, find $$\int_{0}^{\pi} f(x) dx$$

Accepted Answer

Solution:
$$\int_{0}^{1} f(x) dx + \int_{1}^{\pi} f(x) dx$$
$$\int_{0}^{1} x dx + \int_{1}^{\pi} \sin x dx$$
$$= (x)_0^1 + (-\cos x)_1^{\pi} = 2 + \cos 1$$

Question:

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