question 1. 1 point evaluate the following definite integral: (int_{-1}^{0}2x^{2}(4x^{3}-1)^{3}dx) enter…

Explanation:

Step1: Expand the integrand

First, expand ((4x^{3}-1)^{3}) using the formula ((a - b)^{3}=a^{3}-3a^{2}b + 3ab^{2}-b^{3}), where (a = 4x^{3}) and (b = 1). So ((4x^{3}-1)^{3}=(4x^{3})^{3}-3\times(4x^{3})^{2}\times1+3\times4x^{3}\times1^{2}-1^{3}=64x^{9}-48x^{6}+12x^{3}-1). Then the integral (\int_{-1}^{0}2x^{2}(4x^{3}-1)^{3}dx=\int_{-1}^{0}(128x^{11}-96x^{8}+24x^{5}-2x^{2})dx).

Step2: Apply the power - rule for integration

The power - rule for integration is (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C) ((n\neq - 1)). (\int_{-1}^{0}(128x^{11}-96x^{8}+24x^{5}-2x^{2})dx=\left[\frac{128x^{12}}{12}-\frac{96x^{9}}{9}+\frac{24x^{6}}{6}-\frac{2x^{3}}{3}\right]{-1}^{0}). Simplify the antiderivative: (\left[\frac{32x^{12}}{3}-\frac{32x^{9}}{3}+4x^{6}-\frac{2x^{3}}{3}\right]{-1}^{0}).

Step3: Evaluate the definite integral

Substitute the upper and lower limits. When (x = 0), (\frac{32(0)^{12}}{3}-\frac{32(0)^{9}}{3}+4(0)^{6}-\frac{2(0)^{3}}{3}=0). When (x=-1), (\frac{32(-1)^{12}}{3}-\frac{32(-1)^{9}}{3}+4(-1)^{6}-\frac{2(-1)^{3}}{3}=\frac{32}{3}+\frac{32}{3}+4+\frac{2}{3}=\frac{32 + 32+12 + 2}{3}=\frac{78}{3}=26). Then (0-26=-26).

Answer:

(-26)