43. $$ \int_{0}^{\pi} f(x) d x $$ donde $$ f(x)=\left\{\begin{array}{ll}\operatorname{sen} x & ext { si } 0 \leqslant x

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43. $$ \int_{0}^{\pi} f(x) d x $$ donde $$ f(x)=\left\{\begin{array}{ll}\operatorname{sen} x & 	ext { si } 0 \leqslant x<\pi / 2 \ \cos x & 	ext { si } \pi / 2 \leqslant x \leqslant \pi\end{array}ight. $$

UpStudy AI · Accepted Answer

Consideremos la integral definida por la función escalonada

$$
\int_{0}^{\pi} f(x) \, dx,
$$

donde

$$
f(x)=\begin{cases}
\sin x & \text{si } 0 \leq x < \frac{\pi}{2},\[6pt]
\cos x & \text{si } \frac{\pi}{2} \leq x \leq \pi.
\end{cases}
$$

Dividiremos la integral en dos partes, de acuerdo con la definición de $$ f(x) $$:

$$
\int_{0}^{\pi} f(x) \, dx = \int_{0}^{\frac{\pi}{2}} \sin x \, dx + \int_{\frac{\pi}{2}}^{\pi} \cos x \, dx.
$$

1. **Evaluación de** $$\displaystyle \int_{0}^{\frac{\pi}{2}} \sin x \, dx$$:

La integral de $$\sin x$$ es $$-\cos x$$. Evaluamos:

$$
   \int_{0}^{\frac{\pi}{2}} \sin x \, dx = \left[-\cos x \right]_{0}^{\frac{\pi}{2}} = \Bigl[-\cos \frac{\pi}{2}\Bigr] - \Bigl[-\cos 0\Bigr].
   $$

Sabemos que $$\cos \frac{\pi}{2} = 0$$ y $$\cos 0 = 1$$, por lo que:

$$
   \left[-\cos \frac{\pi}{2}\right] - \left[-\cos 0\right] = \bigl[0\bigr] - \bigl[-1\bigr] = 1.
   $$

2. **Evaluación de** $$\displaystyle \int_{\frac{\pi}{2}}^{\pi} \cos x \, dx$$:

La integral de $$\cos x$$ es $$\sin x$$. Evaluamos:

$$
   \int_{\frac{\pi}{2}}^{\pi} \cos x \, dx = \left[\sin x\right]_{\frac{\pi}{2}}^{\pi} = \sin \pi - \sin \frac{\pi}{2}.
   $$

Como $$\sin \pi = 0$$ y $$\sin \frac{\pi}{2} = 1$$, obtenemos:

$$
   0 - 1 = -1.
   $$

3. **Suma de las dos partes:**

$$
   \int_{0}^{\pi} f(x) \, dx = 1 + (-1) = 0.
   $$

Por lo tanto, el valor de la integral es:

$$
\boxed{0}.
$$
La integral de $$ f(x) $$ desde 0 hasta $$ \pi $$ es 0.

donde

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