Evaluate exactly the value of the integral from negative 1 to 0 of the product of the cube of the quantity 2 times x to the 4th power plus 8 times x and 4 times x to the 3rd power plus 4, dx. Your work must include the use of substitution and the antiderivative.

Answer:

-162

Step-by-step explanation:

Integration by substitution is a technique in calculus where a change of variables is made to simplify the integral by expressing it in terms of a new variable, making the integration easier to manage.

To evaluate the following integral, use the method of substitution:

[tex]\displaystyle \int_{-1}^0 (2x^4 + 8x)^3 (4x^3 + 4) \;\text{d}x[/tex]

Let u = 2x⁴ + 8x.

Differentiate u with respect to x using the power rule for differentiation by multiplying each term by its exponent and then subtracting 1 from the exponent:

[tex]\dfrac{\text{d}u}{\text{d}x}=4 \cdot 2x^{4-1}+1\cdot 8 x^{1-1}[/tex]

[tex]\dfrac{\text{d}u}{\text{d}x}=8x^{3}+8[/tex]

Rewrite it so that dx is on its own:

[tex]\text{d}u=8x^{3}+8\;\text{d}x[/tex]

[tex]\text{d}x=\dfrac{1}{8x^{3}+8}\;\text{d}u[/tex]

Change the limits of integration from x to u:

[tex]\begin{aligned}x=-1 \implies u&=2(-1)^4 + 8(-1)\\&=2(1)-8\\&=2-8\\&=-6\end{aligned}[/tex]

[tex]\begin{aligned}x=0 \implies u&=2(0)^4 + 8(0)\\&=2(0)-0\\&=0-0\\&=0\end{aligned}[/tex]

Rewrite the integral in terms of u:

[tex]\begin{aligned}\displaystyle \int_{-1}^0 (2x^4 + 8x)^3 (4x^3 + 4) \;\text{d}x&=\int_{-6}^0 u^3 (4x^3 + 4) \cdot \dfrac{1}{8x^{3}+8}\;\text{d}u\\\\&=\int_{-6}^0 u^3 (4x^3 + 4) \cdot \dfrac{1}{2(4x^{3}+4)}\;\text{d}u\\\\&=\int_{-6}^0\dfrac{u^3}{2} \;\text{d}u\end{aligned}[/tex]

Integrate with respect to u using the power rule for integration by adding 1 to the exponent of each term and then dividing by the new exponent:

[tex]\begin{aligned}\displaystyle \int_{-6}^0\dfrac{u^3}{2} &=\left[\dfrac{u^{3+1}}{2 \cdot (3+1)}\right]^0_{-6}\\\\&=\left[\dfrac{u^{4}}{8}\right]^0_{-6}\\\\&=\dfrac{(0)^4}{8}-\dfrac{(-6)^4}{8}\\\\&=0-\dfrac{1296}{8}\\\\&=-162\end{aligned}[/tex]

Therefore, the value of the given integral is -162.

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