Question

Evaluate the following integrals. $$\int_{1}^{\ln 8} \int_{1}^{\sqrt{z}} \int_{\ln y}^{\ln 2 y} e^{x+y^{2}-z} d x d y d z$$

Short answer

The final answer when we evaluate the triple integral is: $$\frac{1}{2}(\ln 8 - 1)e^{\ln 8 - 1} - \frac{1}{2}(\ln 8)^2 - \frac{1}{2}\ln 8 - \frac{1}{2}\ln 8e^{\ln 8}.$$

Step-by-step solution

Evaluate the inner integral with respect to x

To evaluate the inner integral with respect to x, we have: $$\int_{\ln y}^{\ln 2 y} e^{x+y^{2}-z} dx$$ Integrating this with respect to x, we get: $$e^{x+y^{2}-z} |_{\ln y}^{\ln 2 y} = e^{\ln 2y+y^{2}-z} - e^{\ln y+y^{2}-z}$$ Now, we will continue by evaluating the second integral with respect to y.

Evaluate the second integral with respect to y

After evaluating the first integral, we now have the second integral: $$\int_{1}^{\sqrt{z}} (e^{\ln 2y+y^{2}-z} - e^{\ln y+y^{2}-z}) dy$$ Let's separate the two terms and rewrite the exponents using properties of logarithms: $$\int_{1}^{\sqrt{z}} (2y \cdot e^{y^{2}-z} - y\cdot e^{y^{2}-z}) dy$$ Now, factor out the common term \(ye^{y^{2}-z}\). After that, integrate with respect to y: $$\int_{1}^{\sqrt{z}} (2y-y) ye^{y^{2}-z} dy = \int_{1}^{\sqrt{z}} ye^{y^{2}-z} dy$$ To evaluate this integral, we will use substitution. Let \(u=y^{2}-z\), so \(\frac{1}{2}du=dy\). Then, the limits of integration will be \(u(1)=1-z\) and \(u(\sqrt{z})=z-z=0\). Thus, we have: $$\int_{1-z}^{0} \frac{1}{2}ue^u du$$ Now, to integrate the expression, we will use integration by parts. Let \(v=ue^u\), \(dw=\frac{1}{2}e^u du\), so \(w=\frac{1}{2}e^u\) and \(dv=e^u (1-u)du\). Then we integrate by parts: $$\left(\frac{1}{2}ue^u\right)\Big|_{1-z}^0 - \int_{1-z}^0 \frac{1}{2}e^u (1-u) du$$ Evaluate the first term: $$\frac{1}{2}(0)-(1-z)\frac{1}{2}e^{1-z}$$ Now we integrate the second part: $$ - \int_{1-z}^0 \frac{1}{2}e^u - u\frac{1}{2}e^u du= \left[-\frac{1}{2}e^u + \frac{1}{2}ue^u + \frac{1}{2}e^u\right]_{1-z}^0$$ Substitute back the value of y and combine the two parts to obtain: $$\frac{1}{2}(1-z)e^{1-z} - \frac{1}{2}y^2e^0 + \frac{1}{2}e^0 + \frac{1}{2}ye^0$$ Now, we have the last integral with respect to z.

Evaluate the last integral with respect to z

Lastly, we evaluate the triple integral over z: $$\int_{1}^{\ln 8} \frac{1}{2}(1-z)e^{1-z} - \frac{1}{2}y^2e^0 + \frac{1}{2}e^0 + \frac{1}{2}ye^0 dz$$ Now, integrating all the terms with respect to z, we get the result: $$\left[\frac{1}{2}(1-z)e^{1-z} - \frac{1}{2}z^2y^2 + \frac{1}{2}z + \frac{1}{2}yz\right]_{1}^{\ln 8}$$ Plug in the limits of integration and subtract to find the evaluation of the triple integral: $$\frac{1}{2}(\ln 8 - 1)e^{\ln 8 - 1} - \frac{1}{2}(ln 8)^2 - \frac{1}{2}\ln 8 - \frac{1}{2}\ln 8e^{\ln 8}$$

Key concepts

Integration by Parts

Integration by parts is a technique used to evaluate integrals, especially when they are products of functions that are not easily integrated on their own. The rule is derived from the product rule for differentiation and is given by:

• \( \int u \, dv = uv - \int v \, du \)

To understand this, consider two functions \( u \) and \( dv \). We choose them such that after differentiating \( u \) into \( du \) and integrating \( dv \) to get \( v \), the resulting integral \( \int v \, du \) becomes simpler to solve. In the context of our exercise, integration by parts helps manage the complex exponential expression by focusing on one part of the product.

When applying this technique, it is crucial to choose \( u \) and \( dv \) wisely. Usually, we pick \( u \) to be a function which becomes simpler when differentiated and \( dv \) to be a portion that remains manageable when integrated. Common guidelines include picking inverse trigonometric or logarithmic functions for \( u \) when they are present. However, these decisions often depend on the specific integral at hand.

Substitution Method

The substitution method, also known as \( u \)-substitution, is another powerful technique for solving integrals. This method is particularly useful when the integral contains a composite function that can become simpler with a change of variable. The core idea is to substitute a part of the integral with a single variable, \( u \), and replace the differentials accordingly.

• First, identify a function within the integral that can simplify the integration process.
• We then set \( u \) equal to this function, such as \( u = y^2 - z \) in the initial exercise.
• Determine \( du \) by differentiating \( u \) with respect to the other variables.
• Switch the limits of integration if dealing with definite integrals, as seen in changing limits for \( u(1) = 1-z \) and \( u(\sqrt{z}) = 0 \).

The substitution transforms the integral into a simpler form that is often easier to evaluate. After solving the integral in terms of \( u \), don't forget to substitute back the original variables for the final answer. This method is particularly handy for integrals of exponential, trigonometric, or polynomial functions where direct integration is cumbersome.

Definite Integrals

Definite integrals, as opposed to indefinite integrals, evaluate the accumulation of quantities over a specific interval

• This integration is represented by \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper limits of integration, respectively.
• Definite integrals have a geometric interpretation as the area under the curve of the function \( f(x) \) from \( x = a \) to \( x = b \).

In solving a definite integral, like in the exercise provided, the last step involves substituting back the limits after finding an antiderivative. Evaluate the antiderivative at the upper limit and then subtract the evaluation at the lower limit.

The calculation of definite integrals is crucial for determining the total accumulation of quantities, such as areas, volumes, and total quantities in applied contexts. They are used in physics for calculating work, in economics for finding consumer surplus, and in many other fields that rely on continuous data. Definite integrals provide a tangible conclusion to an integral, representing a specific total quantity rather than a general family of functions.