Calculus: Help ASAP
Evaluate exactly the value of the integral from negative 1 to 0 of the product of the cube of the quantity 4 times x to the 6th power plus 2 times x and 12 times x to the 5th power plus 1, dx. Your work must include the use of substitution and the antiderivative.
Answers
[tex]\bf \textit{now, we'll change the bounds, using u(x)} \\\\\\ u(-1)=4(-1)^6+2(-1)\implies u(-1)=2 \\\\\\ u(0)=4(0)^6+2()\implies u(0)=0\\\\ -------------------------------\\\\ \displaystyle \cfrac{1}{2}\int\limits_{2}^{0}~u^3\cdot du\implies \left. \cfrac{1}{2}\cdot \cfrac{u^4}{4} \right]_{2}^{0}\implies \left. \cfrac{u^4}{8} \right]_{2}^{0}\implies [0]-[2]\implies -2[/tex]
Answer:
2.264 (3 d.p.)
Step-by-step explanation:
Given integral:
[tex]\displaystyle \int^0_{-1} \left(4x^6+2x\right)^3\left(12x^5+1\right)\;\text{d}x[/tex]
First, evaluate the indefinite integral using the method of substitution.
[tex]\textsf{Let} \;\;u = 4x^6+2x[/tex]
Find du/dx and rewrite it so that dx is on its own:
[tex]\dfrac{\text{d}u}{\text{d}x}=24x^5+2 \implies \text{d}x=\dfrac{1}{24x^5+2}\; \text{d}u[/tex]
Rewrite the original integral in terms of u and du, and evaluate:
[tex]\begin{aligned}\displaystyle \int\left(4x^6+2x\right)^3\left(12x^5+1\right)\;\text{d}x&=\int \left(u\right)^3\left(12x^5+1\right)\cdot \dfrac{1}{24x^5+2}\; \text{d}u\\\\&=\int \left(u\right)^3\left(12x^5+1\right)\cdot \dfrac{1}{2(12x^5+1)}\; \text{d}u\\\\&=\int \dfrac{u^3\left(12x^5+1\right)}{2(12x^5+1)}\; \text{d}u\\\\&=\displaystyle \int \dfrac{u^3}{2}\; \text{d}u\\\\&=\dfrac{u^{3+1}}{2(3+1)}+C\\\\&=\dfrac{u^4}{8}+C\end{aligned}[/tex]
Substitute back u = 4x⁶ + 2x:
[tex]=\dfrac{(4x^6+2x)^4}{8}+C[/tex]
Therefore:
[tex]\displaystyle \int \left(4x^6+2x\right)^3\left(12x^5+1\right)\;\text{d}x=\dfrac{(4x^6+2x)^4}{8}+C[/tex]
To evaluate the definite integral, we must first determine any intervals within the given interval -1 ≤ x ≤ 0 where the curve lies below the x-axis. This is because when we integrate a function that lies below the x-axis, it will give a negative area value.
Find the x-intercepts by setting the function to zero and solving for x.
[tex]\left(4x^6+2x\right)^3\left(12x^5+1\right)=0[/tex]
Therefore:
[tex]\begin{aligned}\left(4x^6+2x\right)^3&=0\\4x^6+2x&=0\\x(4x^5+2)&=0\end{aligned}[/tex]
[tex]x=0[/tex]
[tex]\begin{aligned}4x^5+2&=0\\4x^5&=-2\\x^5&=-\frac{1}{2}\\x&=\sqrt[5]{-\dfrac{1}{2}}\end{aligned}[/tex]
[tex]\begin{aligned}12x^5+1&=0\\12x^5&=-1\\x^5&=-\dfrac{1}{12}\\x&=\sqrt[5]{-\dfrac{1}{12}}\end{aligned}[/tex]
Therefore, the curve of the function is:
Below the x-axis between -1 and ⁵√(-1/2).Above the x-axis between ⁵√(-1/2) and ⁵√(-1/12).Below the x-axis between ⁵√(-1/12) and 0.So to calculate the total area, we need to calculate the positive and negative areas separately and then add them together, remembering that if you integrate a function to find an area that lies below the x-axis, it will give a negative value.
Integrate the function between -1 and ⁵√(-1/2).
As the area is below the x-axis, we need to negate the integral so that the resulting area is positive:
[tex]\begin{aligned}A_1&=-\displaystyle \int_{-1}^{\sqrt[5]{-\frac{1}{2}}} \left(4x^6+2x\right)^3\left(12x^5+1\right)\;\text{d}x\\\\&=-\left[\dfrac{(4x^6+2x)^4}{8}\right]_{-1}^{\sqrt[5]{-\frac{1}{2}}}\\\\&=-\left[\left(\dfrac{\left(4\left(\sqrt[5]{-\frac{1}{2}}\right)^6+2\left(\sqrt[5]{-\frac{1}{2}}\right)\right)^4}{8}\right)-\left(\dfrac{(4(-1)^6+2(-1))^4}{8}\right)\right]\\\\&=-[0-2]\\\\&=2\end{aligned}[/tex]
Integrate the function between ⁵√(-1/2) and ⁵√(-1/12).
[tex]\begin{aligned}A_2&=\displaystyle \int_{\sqrt[5]{-\frac{1}{2}}} ^{\sqrt[5]{-\frac{1}{12}}} \left(4x^6+2x\right)^3\left(12x^5+1\right)\;\text{d}x\\\\&=\left[\dfrac{(4x^6+2x)^4}{8}\right]_{\sqrt[5]{-\frac{1}{2}}}^{\sqrt[5]{-\frac{1}{12}}}\\\\&=\left(\dfrac{\left(4\left(\sqrt[5]{-\frac{1}{12}}\right)^6+2\left(\sqrt[5]{-\frac{1}{12}}\right)\right)^4}{8}\right)-\left(\dfrac{\left(4\left(\sqrt[5]{-\frac{1}{2}}\right)^6+2\left(\sqrt[5]{-\frac{1}{2}}\right)\right)^4}{8}\right)\\\\\end{aligned}[/tex]
[tex]\begin{aligned}&=\dfrac{625}{648\sqrt[5]{12^4}}-0\\\\&=0.132117398...\end{aligned}[/tex]
Integrate the function between ⁵√(-1/12) and 0.
As the area is below the x-axis, we need to negate the integral so that the resulting area is positive:
[tex]\begin{aligned}A_3&=-\displaystyle \int_{\sqrt[5]{-\frac{1}{12}}}^0 \left(4x^6+2x\right)^3\left(12x^5+1\right)\;\text{d}x\\\\&=-\left[\dfrac{(4x^6+2x)^4}{8}\right]_{\sqrt[5]{-\frac{1}{12}}}^0\\\\&=-\left[\left(\dfrac{(4(0)^6+2(0))^4}{8}\right)-\left(\dfrac{\left(4\left(\sqrt[5]{-\frac{1}{12}}\right)^6+2\left(\sqrt[5]{-\frac{1}{12}}\right)\right)^4}{8}\right)\right]\\\\&=-\left[0-\dfrac{625}{648\sqrt[5]{12^4}}\right]\\\\&=\dfrac{625}{648\sqrt[5]{12^4}}\\\\&=0.132117398...\\\\\end{aligned}[/tex]
To evaluate the definite integral, sum A₁, A₂ and A₃:
[tex]\begin{aligned}\displaystyle \int^0_{-1} \left(4x^6+2x\right)^3\left(12x^5+1\right)\;\text{d}x&=2+2\left( \dfrac{625}{648\sqrt[5]{12^4}}\right)\\\\&=2+ \dfrac{625}{324\sqrt[5]{12^4}}\right}\\\\&=2.264\; \sf (3\;d.p.)\end{aligned}[/tex]
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