The given image presents a triple integral:

$\int_{0}^{\pi/4} \int_{0}^{\ln \sec z} \int_{-\infty}^{2y} e^x \, dx \, dy \, dz$

Let’s evaluate this step-by-step:

Step 1: Evaluate the innermost integral

The innermost integral is:

$\int_{-\infty}^{2y} e^x \, dx$

The antiderivative of $e^x$ is $e^x$. Thus, we have:

$\int_{-\infty}^{2y} e^x \, dx = \left[ e^x \right]_{-\infty}^{2y}$

At $x = -\infty$, $e^x \to 0$. At $x = 2y$, $e^x = e^{2y}$. Therefore:

$\int_{-\infty}^{2y} e^x \, dx = e^{2y} - 0 = e^{2y}$

Step 2: Evaluate the second integral

The second integral is:

$\int_{0}^{\ln \sec z} e^{2y} \, dy$

Since $e^{2y}$ depends only on $y$, let’s compute the antiderivative of $e^{2y}$. The antiderivative is:

$\int e^{2y} \, dy = \frac{1}{2} e^{2y}$

Now evaluate the definite integral:

$\int_{0}^{\ln \sec z} e^{2y} \, dy = \left[ \frac{1}{2} e^{2y} \right]_{0}^{\ln \sec z}$

At $y = 0$, $e^{2y} = e^0 = 1$. At $y = \ln \sec z$, $e^{2y} = e^{2 \ln \sec z} = (\sec z)^2$. Therefore:

$\int_{0}^{\ln \sec z} e^{2y} \, dy = \frac{1}{2} (\sec z)^2 - \frac{1}{2}(1) = \frac{1}{2} ((\sec z)^2 - 1)$

Using the trigonometric identity $(\sec z)^2 - 1 = (\tan z)^2$, this simplifies to:

$\int_{0}^{\ln \sec z} e^{2y} \, dy = \frac{1}{2} (\tan z)^2$

Step 3: Evaluate the outermost integral

The outermost integral is:

$\int_{0}^{\pi/4} \frac{1}{2} (\tan z)^2 \, dz$

The antiderivative of $(\tan z)^2$ is $\tan z - z$. Thus:

$\int (\tan z)^2 \, dz = \tan z - z$

Now evaluate the definite integral:

$\int_{0}^{\pi/4} (\tan z)^2 \, dz = \left[ \tan z - z \right]_{0}^{\pi/4}$

At $z = \pi/4$, $\tan z = 1$, and $z = \pi/4$. At $z = 0$, $\tan z = 0$, and $z = 0$. Therefore:

$\int_{0}^{\pi/4} (\tan z)^2 \, dz = \left( 1 - \frac{\pi}{4} \right) - \left( 0 - 0 \right) = 1 - \frac{\pi}{4}$

Thus:

$\int_{0}^{\pi/4} \frac{1}{2} (\tan z)^2 \, dz = \frac{1}{2} \left( 1 - \frac{\pi}{4} \right)$

Final Answer:

The value of the triple integral is:

$\boxed{\frac{1}{2} \left( 1 - \frac{\pi}{4} \right)}$

Let me know if you need any further clarifications or additional steps explained!

Related Questions:

1. What is the significance of changing the order of integration in triple integrals?
2. How do you evaluate improper integrals involving limits of infinity?
3. What are the properties of exponential functions in multiple integrals?
4. How does the trigonometric identity $(\sec z)^2 - 1 = (\tan z)^2$ simplify integration?
5. What are the common applications of triple integrals in physics and engineering?

Tip:

For problems involving multiple integrals, always evaluate from the innermost to the outermost integral to simplify the process systematically.