Question

Evaluate the following integrals. $$\int_{1}^{\sqrt[3]{2}} y^{8} e^{y^{3}} d y$$

Short answer

Answer: The substitution made initially is u = y^3.

Step-by-step solution

Perform the substitution

Let \(u = y^3\). Therefore, \(du = 3y^2 dy\). We can write \(dy\) in terms of \(du\) like this: $$dy = \frac{du}{3y^2}$$ Now let's find the new limits of integration in terms of \(u\). When \(y = 1\), \(u = 1^3 = 1\). When \(y = \sqrt[3]{2}\), \(u = (\sqrt[3]{2})^3 = 2\). So the new limits of integration will be \(1\) to \(2\). Now, we'll replace \(y^3\) with \(u\), and \(dy\) with \(\frac{du}{3y^2}\): $$\int_{1}^{2} y^8 e^u \frac{du}{3y^2}$$

Simplify the integrand

The integrand that we have now is: $$y^8 e^u \frac{du}{3y^2}$$ Simplify the expression by cancelling out common terms: $$\frac{y^6 e^u}{3} du$$ Now, let's express \(y^6\) in terms of \(u\). Since we know that \(u = y^3\), we can write \(y^6\) as: $$y^6 = (y^3)^2 = u^2$$ Substitute this back into the expression: $$\int_1^2 \frac{u^2 e^u}{3} du$$

Integrate and apply the limits of integration

Now we integrate the function with respect to \(u\): $$\int_1^2 \frac{u^2 e^u}{3} du = \frac{1}{3} \int_1^2 u^2 e^u du$$ To integrate the above expression, we need to use integration by parts twice. Let \(v = u^2\) and \(dw = e^u du\). Differentiate \(v\) to find \(dv = 2u du\) and integrate \(dw\) to find \(w = e^u\). Applying integration by parts formula: \(\int v dw = vw - \int w dv\) $$\frac{1}{3}(u^2 e^u - 2 \int u e^u du)$$ Now, we need to use integration by parts for the second integral: Let \(s = u\) and \(dx = e^u du\). Differentiate \(s\) to find \(ds = du\) and integrate \(dx\) to find \(x = e^u\). $$\frac{1}{3}[u^2 e^u - 2(ux - \int x ds)]$$ Now let's integrate and apply the limits of integration from 1 to 2: $$\frac{1}{3}[2^2 e^2 - 1^2 e^1 - 2(2e^2 - e^1 - (e^2 - e^1))]$$

Simplify and find the final answer

Now let's simplify the expression and find the final value of the integral: $$\frac{1}{3}[4e^2 - e - 4e^2 + 2e + 2(e^2 - e)]$$ $$\frac{1}{3}[(4e^2 - 4e^2 + 2e^2) - (e - 2e + 2e)]$$ $$\frac{1}{3}[2e^2]$$ The definite integral evaluates to: $$\frac{2e^2}{3}$$

Key concepts

Integration by Parts

Integration by parts is a useful technique for solving complex integrals. It helps break down a product of functions into simpler parts. The formula for integration by parts is:\[ \int v \, dw = vw - \int w \, dv \]In this formula, we typically choose one part of the integrand to differentiate (\(v\)) and the other part to integrate (\(dw\)). In the exercise above, during the integration process, integration by parts is applied twice to solve \(\int u^2 e^u \, du\).

Let's understand the steps:

• First, set \(v = u^2\) and \(dw = e^u \, du\). Compute \(dv = 2u \, du\) and \(w = e^u\).
• Apply the integration by parts formula: \(vw - \int w \, dv\).
• For the new integral \(\int u \, e^u \, du\), repeat the integration by parts process by choosing \(s = u\) and \(dx = e^u \, du\).
• Compute \(ds = du\) and \(x = e^u\), then apply the formula again.

Using integration by parts iteratively helps solve more complicated integrals step by step, as seen in the exercise solution.

Substitution Method

The substitution method is a technique used to simplify integration by changing variables. This helps to transform a complicated integral into a simpler one. In the given exercise, the substitution method is used at the beginning to make the integral more manageable. Let's break it down:

Here's how it works in the context of the problem:

• Choose a substitution that simplifies the integrand. Here, \(u = y^3\) was chosen.
• Differentiate \(u\) to get \(du = 3y^2 \, dy\) and solve for \(dy\) to express it in terms of \(du\) and \(y\).
• Update the limits of integration based on the variable change: for \(y = 1\), \(u = 1\), and for \(y = \sqrt[3]{2}\), \(u = 2\).
• Replace \(y^3\) with \(u\) and express \(y^6\) as \(u^2\) because \(y^6 = (y^3)^2 = u^2\).
• This transformation makes it easier to work with the integral as \((1/3)\int u^2 e^u \, du\).

Using substitution, complex integrals are reduced to simpler forms, making them solvable by other integration techniques thereafter.

Definite Integrals

Definite integrals are a core concept in calculus, representing the area under a curve between two points. A definite integral is evaluated from specific bounds, providing a numerical value rather than a general function like an indefinite integral. In this exercise, the integral \(\int_1^{\sqrt[3]{2}} y^8 e^{y^3} \, dy\) has specified bounds from 1 to \(\sqrt[3]{2}\).

Key features of definite integrals include:

• The limits of integration are transformed during substitution, as seen when \(y = 1\) translates to \(u = 1\) and \(y = \sqrt[3]{2}\) becomes \(u = 2\).
• After integrating, the result is evaluated using these limits, which was done by calculating \([(2^2 e^2 - e^1) - (other terms)]\) in the example.
• The evaluation process involves substituting the limits back into the integrated function and simplifying the result to find the area. That led to the result \(\frac{2e^2}{3}\).

Definite integrals effectively provide a finite value that can represent various real-world quantities like areas, distances, or accumulated quantities, depending on the context.