92–98. Evaluate the following integrals.
92. ∫[1 to √2] y⁸ e^(y²) dy

Verified step by step guidance

1. Identify the integral to solve: \(\int_{1}^{\sqrt{2}} y^{8} e^{y^{2}} \, dy\).
2. Consider a substitution to simplify the integral. Let \(u = y^{2}\), which implies \(du = 2y \, dy\) or equivalently \(y \, dy = \frac{du}{2}\).
3. Rewrite the integral in terms of \(u\). Express \(y^{8}\) as \((y^{2})^{4} = u^{4}\), and replace \(y \, dy\) with \(\frac{du}{2}\). The integral becomes \(\int_{u=1^{2}}^{u=(\sqrt{2})^{2}} u^{4} e^{u} \cdot y^{7} \, dy\). However, since \(y^{7} \, dy\) is not directly replaced by \(du\), we need to express \(y^{8} \, dy\) carefully.
4. Notice that \(y^{8} \, dy = y^{7} \cdot y \, dy\). Using the substitution, \(y \, dy = \frac{du}{2}\), so \(y^{8} \, dy = y^{7} \cdot \frac{du}{2}\). But \(y^{7} = (y^{2})^{3} \cdot y = u^{3} \cdot y\). This suggests the substitution might be more complex, so consider an alternative approach such as integration by parts.
5. Set up integration by parts with \(u = y^{7}\) and \(dv = y e^{y^{2}} \, dy\). Then compute \(du\) and \(v\), and apply the integration by parts formula \(\int u \, dv = uv - \int v \, du\). This will help break down the integral into simpler parts.

Key Concepts

Integration by Substitution

Integration by substitution is a method used to simplify integrals by changing variables. It involves identifying a part of the integrand whose derivative also appears in the integral, allowing the integral to be rewritten in terms of a new variable. This technique is especially useful when dealing with composite functions like e^(y²).

Exponential Functions in Integration

Exponential functions, such as e^(y²), often require special techniques for integration because their antiderivatives are not elementary. Recognizing the structure of the exponent and how it relates to other parts of the integrand is crucial for applying substitution or integration by parts effectively.

Definite Integrals and Limits of Integration

Definite integrals calculate the accumulated area under a curve between two specific points. Understanding how to apply the limits of integration after finding the antiderivative is essential to evaluate the integral's exact value. Proper substitution also requires adjusting these limits to match the new variable.