Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
The area is 0.25
What is a tangent line?
The straight line that "just touches" the curve at a particular location is known as the tangent line (or simply tangent) to a plane curve in geometry. It was described by Leibniz as the path connecting two points on a curve that are infinitely near together. A straight line has a slope of f'(c), where f' is the derivative of f, and is said to be tangent to a curve at a point x = c if it passes through the point (c, f(c)) on the curve. Space curves and curves in n-dimensional Euclidean space have a similar definition.
[tex]$\begin{aligned}& f^{\prime}(x)=\frac{\mathrm{d}}{\mathrm{d} x}\left[x^3-2 x\right] \\& f^{\prime}(x)=\frac{\mathrm{d}}{\mathrm{d} x}\left[x^3\right]-\frac{\mathrm{d}}{\mathrm{d} x}[2 x] \quad \text { Sum } / \text { Rest rule } \\& f^{\prime}(x)=3 x^{3-1}-2 \quad \text { Power rule } \\& f^{\prime}(x)=3 x^2-2 \\&\end{aligned}$[/tex]
Now finding tangent for a=-1,
[tex]\begin{aligned}& y=f^{\prime}(a)(x-a)+f(a) \\& y=f^{\prime}(-1)(x+1)+f(-1) \\& y=\left(3(-1)^2-2\right)(x+1)+(-1)^3-2(-1) \\& \mathbf{y}=\mathbf{x}+\mathbf{2}\end{aligned}$[/tex]
For the first area,
[tex]$\begin{aligned}& A_1: \int_{-2}^{-\sqrt{2}} x+2 d x=\frac{1}{2} x^2+\left.2 x\right|_{-2} ^{-\sqrt{2}} \\& A_1=\frac{1}{2}(-\sqrt{2})^2+2(-\sqrt{2})-\left(\frac{1}{2}(-2)^2+2(-2)\right) \\& A_1=\mathbf{3}-\mathbf{2} \sqrt{\mathbf{2}} \text { squared units } \\& \mathbf{A}_{\mathbf{1}} \approx \mathbf{0 . 1 7 1 5 7 2 8 7 6} \ldots\end{aligned}$[/tex]
For the second area,
[tex]$\begin{aligned}& A_2: \int_{-\sqrt{2}}^{-1} x+2-\left(x^3-2 x\right) d x=-\frac{1}{4} x^4+\frac{3}{2} x^2+\left.2 x\right|_{-\sqrt{2}} ^{-1} \\& A_2=-\frac{1}{4}(-1)^4+\frac{3}{2}(-1)^2+2(-1)-\left(-\frac{1}{4}(-\sqrt{2})^4+\frac{3}{2}(-\sqrt{2})^2+2(-\sqrt{2})\right) \\& A_2=-\frac{\mathbf{1 1}}{\mathbf{4}}+\mathbf{2} \sqrt{\mathbf{2}} \text { squared units } \\& \mathbf{A}_{\mathbf{2}} \approx \mathbf{0 . 0 7 8 4 2 7 1 2 4} \ldots .\end{aligned}$[/tex]
Hence, required area is
[tex]$\begin{aligned}& A=A_1+A_2 \\& A=3-2 \sqrt{2}-\frac{11}{4}+2 \sqrt{2} \\& \mathbf{A}=\frac{\mathbf{1}}{\mathbf{4}}=\mathbf{0 . 2 5}\end{aligned}$[/tex]
The area is 0.25
To know more about tangent visit:
brainly.com/question/10053881
#SPJ4
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
