i)
Z 18
1
r
3
z
dz
j)
Z 1
0
(x
e + e
x
) dx
k)
Z √
3
1/
√
3
8
1 + x
2
dx
l)
Z 1/
√
2
1/2
dx
√
1 − x
2
m)
Z π
0
f(x) dx, donde f(x) =



sin x, 0 ≤ x ≤
π
2
cos x,
π
2
< x ≤ π.

Let's break down each part of your question step by step and solve the integrals provided.

### i) \(\int_1^3 z \, dz\)
This is a simple integral of a linear function.

\[
\int z \, dz = \frac{z^2}{2} + C
\]

Evaluating from 1 to 3:

\[
\left[\frac{z^2}{2}\right]_1^3 = \frac{3^2}{2} - \frac{1^2}{2} = \frac{9}{2} - \frac{1}{2} = \frac{8}{2} = 4
\]

So, the result is \(4\).

### j) \(\int_0^1 \left(x e + e^x\right) \, dx\)

We will integrate term by term:

\[
\int_0^1 x e \, dx + \int_0^1 e^x \, dx
\]

1. For the first term:
   \[
   \int_0^1 x e \, dx = e \int_0^1 x \, dx = e \left[\frac{x^2}{2}\right]_0^1 = e \times \frac{1}{2} = \frac{e}{2}
   \]

2. For the second term:
   \[
   \int_0^1 e^x \, dx = \left[e^x\right]_0^1 = e^1 - e^0 = e - 1
   \]

Combining both terms:

\[
\frac{e}{2} + (e - 1) = \frac{e}{2} + e - 1 = \frac{3e}{2} - 1
\]

Thus, the result is \(\frac{3e}{2} - 1\).

### k) \(\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{8}{1 + x^2} \, dx\)

This is an integral involving the inverse tangent function, since:

\[
\int \frac{1}{1 + x^2} \, dx = \tan^{-1}(x)
\]

So, the integral becomes:

\[
8 \left[\tan^{-1}(x)\right]_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}
\]

We now evaluate the limits:

1. \(\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}\)
2. \(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\)

Thus:

\[
8 \left(\frac{\pi}{3} - \frac{\pi}{6}\right) = 8 \times \frac{\pi}{6} = \frac{4\pi}{3}
\]

### l) \(\int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{1}{\sqrt{1 - x^2}} \, dx\)

This is the integral of the arcsine function:

\[
\int \frac{1}{\sqrt{1 - x^2}} \, dx = \sin^{-1}(x)
\]

Evaluating at the bounds:

1. \(\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}\)
2. \(\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}\)

Thus, the result is:

\[
\frac{\pi}{4} - \frac{\pi}{6} = \frac{3\pi}{12} - \frac{2\pi}{12} = \frac{\pi}{12}
\]

### m) \(\int_0^\pi f(x) \, dx\), where \( f(x) = \begin{cases} \sin(x) & \text{if } 0 \leq x \leq \frac{\pi}{2} \\ \cos(x) & \text{if } \frac{\pi}{2} < x \leq \pi \end{cases}\)

We will split the integral into two parts:

\[
\int_0^\pi f(x) \, dx = \int_0^{\frac{\pi}{2}} \sin(x) \, dx + \int_{\frac{\pi}{2}}^\pi \cos(x) \, dx
\]

1. For the first integral:
   \[
   \int_0^{\frac{\pi}{2}} \sin(x) \, dx = \left[-\cos(x)\right]_0^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) + \cos(0) = 0 + 1 = 1
   \]

2. For the second integral:
   \[
   \int_{\frac{\pi}{2}}^\pi \cos(x) \, dx = \left[\sin(x)\right]_{\frac{\pi}{2}}^\pi = \sin(\pi) - \sin\left(\frac{\pi}{2}\right) = 0 - 1 = -1
   \]

Thus, the total result is:

\[
1 + (-1) = 0
\]

---

Let me know if you'd like more details or further explanations!

Here are five related questions:
1. How do you compute definite integrals involving trigonometric functions?
2. What is the role of the inverse tangent in integration?
3. How does substitution simplify trigonometric integrals?
4. What are common techniques for integrating rational functions?
5. How do piecewise functions affect the integration process?

**Tip:** Always split the integral based on the function’s domain, especially for piecewise functions.

Z 18 1 r 3 z dz j) Z 1 0 (x e + e^x) dx k) Z √3 1/√3 8 1 + x^2 dx l) Z 1/√2 1/2 dx √1 − x^2 m) Z π 0 f(x) dx, donde f(x) = sin x, 0 ≤ x ≤ π/2; cos x, π/2 < x ≤ π.

Explore detailed solutions for a set of definite integrals involving trigonometric and inverse trigonometric functions. Learn how to handle piecewise functions, and compute integrals like ∫ (1 + x^2)^(-1) dx and ∫ sin(x) dx. Suitable for university-level calculus students.

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