Using properties of integrals Use the value of the first integral I to evaluate the two given integrals. 
I = ∫₀¹ (𝓍³ ― 2𝓍) d𝓍 = ―3/4
(a) ∫₀¹ (4𝓍―2𝓍³) d𝓍

Verified step by step guidance

1. Step 1: Recognize that the given integral in part (a), ∫₀¹ (4𝓍 ― 2𝓍³) d𝓍, can be split into two separate integrals using the linearity property of integrals: ∫₀¹ (4𝓍 ― 2𝓍³) d𝓍 = ∫₀¹ 4𝓍 d𝓍 ― ∫₀¹ 2𝓍³ d𝓍.
2. Step 2: Factor out constants from each integral. For the first term, ∫₀¹ 4𝓍 d𝓍 becomes 4∫₀¹ 𝓍 d𝓍. For the second term, ∫₀¹ 2𝓍³ d𝓍 becomes 2∫₀¹ 𝓍³ d𝓍.
3. Step 3: Use the given value of I = ∫₀¹ (𝓍³ ― 2𝓍) d𝓍 = ―3/4 to extract the value of ∫₀¹ 𝓍³ d𝓍. Rewrite I as ∫₀¹ 𝓍³ d𝓍 ― ∫₀¹ 2𝓍 d𝓍 = ―3/4. This equation can be solved to find ∫₀¹ 𝓍³ d𝓍.
4. Step 4: Compute ∫₀¹ 𝓍 d𝓍 using the power rule for integration. The integral of 𝓍 with respect to 𝓍 is (𝓍²)/2. Evaluate this from 0 to 1 to find the value of ∫₀¹ 𝓍 d𝓍.
5. Step 5: Substitute the values of ∫₀¹ 𝓍³ d𝓍 and ∫₀¹ 𝓍 d𝓍 into the expression for ∫₀¹ (4𝓍 ― 2𝓍³) d𝓍 = 4∫₀¹ 𝓍 d𝓍 ― 2∫₀¹ 𝓍³ d𝓍 to compute the final result.

Key Concepts

Properties of Integrals

The properties of integrals, such as linearity and the ability to split integrals, are fundamental in calculus. Linearity allows us to factor constants out of integrals and combine integrals of the same limits. This means that if we have an integral of a sum, we can separate it into the sum of integrals, which simplifies calculations.

Definite Integrals

Definite integrals represent the signed area under a curve between two points on the x-axis. The notation ∫ₐᵇ f(x) dx indicates the integral of the function f(x) from a to b. The value of a definite integral can be interpreted as the accumulation of quantities, which is essential for evaluating integrals over specific intervals.

Integration Techniques

Various techniques exist for evaluating integrals, including substitution and integration by parts. In this context, recognizing patterns in the integrand can help simplify the integral. For example, if an integral can be expressed in terms of known integrals, such as the one provided (I), it can be evaluated more easily by leveraging previously calculated values.