Question

Evaluate the following iterated integrals. $$\int_{1}^{4} \int_{0}^{2} e^{\sqrt{x}} d y d x$$

Short answer

Question: Evaluate the iterated integral $$\int_{1}^{4} \int_{0}^{2} e^{\sqrt{x}} dy dx$$. Answer: The iterated integral evaluates to $$8e^2$$.

Step-by-step solution

Integrate with respect to \(y\)

First, integrate the function \(e^{\sqrt{x}}\) with respect to \(y\). As \(e^{\sqrt{x}}\) is independent of \(y\), its integral will be a linear function of \(y\). Performing the integration: $$\int_{0}^{2} e^{\sqrt{x}} dy = e^{\sqrt{x}}(y)\Big|_0^2 = 2e^{\sqrt{x}}$$

Integrate with respect to \(x\)

Now, integrate the result \(2e^{\sqrt{x}}\) with respect to \(x\) from 1 to 4: $$\int_{1}^{4} 2e^{\sqrt{x}} dx$$ To evaluate this integral, we perform a change of variables/substitution. Let \(u = \sqrt{x}\), so \(x = u^2\) and \(dx = 2u du\). The limits of integration for \(u\) will be \(\sqrt{1} = 1\) and \(\sqrt{4} = 2\). Then, the integral becomes: $$\int_{1}^{2} 2e^u (2u) du = 4\int_{1}^{2} u e^u du$$

Evaluate the integral using integration by parts

To evaluate this integral, we use integration by parts, where we let \(dv = ue^u du\) and \(v = \int ue^u du\). We then perform another round of integration by parts on \(v\), letting \(dw = u du\) and \(w = \int u du\). Evaluating these integrals gives \(w = \frac{1}{2}u^2\) and \(dw = u du\). Now, for the second round of integration by parts, we have \(v = \int e^u dw = e^u w - \int e^u w' du = e^u (\frac{1}{2}u^2) - \int e^u u du\). Thus, $$v = e^u (\frac{1}{2}u^2) - e^u u + \int e^u du = e^u (\frac{1}{2}u^2 - u + 1)$$ Then, the original integral becomes: $$4\int_{1}^{2} u e^u du = 4 \left[e^u (\frac{1}{2}u^2 - u + 1)\Big|_1^2\right]$$

Evaluate the result and simplify

Now we can substitute the limits of integration for \(u\) and evaluate the result: $$4 \left[e^2 (\frac{1}{2}(2)^2 - 2 + 1) - e^1 (\frac{1}{2}(1)^2 - 1 + 1)\right] = 4 \left[e^2 (2) - e^1 (0)\right] = 8e^2$$ So, the iterated integral evaluates to: $$\int_{1}^{4} \int_{0}^{2} e^{\sqrt{x}} dy dx = 8e^2$$

Key concepts

Integration by Substitution

Integration by substitution is a powerful technique that helps simplify integrals by making a clever change of variables. In our example, the substitution was crucial in handling the integral \( \int_{1}^{4} 2e^{\sqrt{x}} dx \).

Here's a step-by-step breakdown of how this works:

• Identify a part of the integral to substitute. In this case, we used the square root of \(x\), so let's set \(u = \sqrt{x}\).
• Determine how \(dx\) changes in terms of \(du\). Here, \(x = u^2\) implies \(dx = 2u \, du\).
• Replace the original variable with the new one, transforming the limits of integration to match \(u\). The original limits \(x = 1\) and \(x = 4\) become \(u = 1\) and \(u = 2\).
• Substitute these into the equation, transforming the original expression into \(4 \int_{1}^{2} u e^u \, du\).

Now, the integral is simpler to evaluate, as seen in the subsequent steps using integration by parts. This underscores how substitution can make complex integrals more manageable.

Integration by Parts

Integration by parts is another core technique, especially effective for integrals involving products of functions, like \(u e^u\) in this exercise.
It is derived from the product rule for differentiation and follows the formula:

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

In this problem, integration by parts was used twice:

• First, to tackle \(\int u e^u \, du\), by choosing \(v = e^u\) and \(dw = u \, du\).
• This required another integration by parts within it, splitting \(\int ue^u du\) further.
• Finally, it simplified to \(v = e^u (\frac{1}{2}u^2 - u + 1)\).

By understanding integration by parts, you can tackle many problems involving products of polynomials and exponential functions, helping to break down the problem into more digestible pieces.

Exponential Functions

Exponential functions, like \(e^{\sqrt{x}}\), often appear in integrals and can sometimes complicate evaluations. Understanding the properties of exponential functions is essential in integrating expressions involving them.

Key properties include:

• The exponential function \(e^x\) is its own derivative and integral, making it unique.
• Exponential functions grow rapidly, which can affect the convergence and evaluation of integrals.

In the given problem:

• The function \(e^{\sqrt{x}}\) required substitution to change \(\sqrt{x}\) to a simpler variable \(u\).
• Once inside the integral, understanding \(e^u\) allowed it to be manipulated using integration by parts, ultimately facilitating the evaluation.

A solid grasp of exponential functions is invaluable when dealing with complex integrals across calculus, helping make challenging math problems more approachable.