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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
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