Evaluating integrals Evaluate the following integrals.

∫ y² (3y³ + 1)⁴ dy

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Step 1: Recognize that the integral involves a composite function. The term (3y³ + 1)⁴ suggests that substitution might simplify the integral. Let u = 3y³ + 1.

Step 2: Compute the derivative of u with respect to y. Since u = 3y³ + 1, we find that du/dy = 9y². Rearrange to express dy in terms of du: dy = du / 9y².

Step 3: Substitute u and dy into the integral. Replace (3y³ + 1)⁴ with u⁴ and y² dy with y² * (du / 9y²). The y² terms cancel out, leaving the integral as (1/9) ∫ u⁴ du.

Step 4: Integrate u⁴ with respect to u. The integral of u⁴ is (u⁵ / 5). Multiply by the constant (1/9) to get (1/45) u⁵.

Step 5: Substitute back u = 3y³ + 1 into the result to return to the original variable. The final expression is (1/45) (3y³ + 1)⁵ + C, where C is the constant of integration.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Integration

Integration is a fundamental concept in calculus that involves finding the integral of a function, which represents the area under the curve of that function on a given interval. It can be thought of as the reverse process of differentiation. The integral can be definite, providing a numerical value over a specific interval, or indefinite, resulting in a general form of antiderivatives.

Substitution Method

The substitution method is a technique used in integration to simplify the process by changing the variable of integration. This involves substituting a part of the integrand with a new variable, which can make the integral easier to evaluate. It is particularly useful when dealing with composite functions, allowing for a more straightforward integration process.

Polynomial Functions

Polynomial functions are expressions that consist of variables raised to non-negative integer powers and their coefficients. In the context of integration, recognizing polynomial forms is crucial, as they can be integrated using standard rules. The integral of a polynomial function involves increasing the exponent by one and dividing by the new exponent, which is essential for evaluating integrals like the one presented.

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