Question

C ,ex cu volumul corpului

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Answer 1

$\displaystyle 2.~~f:\mathbb{R}\rightarrow \mathbb{R},~f(x)=x^2+3x\\\\ c)~g:[1,2]\rightarrow \mathbb{R},~g(x)=\frac{3f(x)}{x} \\\\ g(x)=\frac{3f(x)}{x} =\frac{3\Big(x^2+3x\Big)}{x}=\frac{3x^2+9x}{x} =\frac{\not x(3x+9)}{\not x} =3x+9\\\\ V=\pi \int\limits^b_a g^2(x)dx=\pi \int\limits^2_1 (3x+9)^2dx=\pi \int\limits^2_1\Big((3x)^2+2 \cdot 3x \cdot 9+9^2\Big)dx=\\\\=\pi \int\limits^2_1\Big(9x^2+54x+81\Big)dx=\pi \Bigg(\int\limits^2_19x^2dx+\int\limits^2_154xdx+\int\limits^2_181dx\Bigg)=$

$\displaystyle =\pi \Bigg(9\int\limits^2_1x^2dx+54\int\limits^2_1xdx+81\int\limits^2_11dx\Bigg)=\pi\Bigg(9 \cdot \frac{x^3}{3} \Bigg|_1^2+54 \cdot \frac{x^2}{2} \Bigg|_1^2+81x\Bigg|_1^2\Bigg)=$

$\displaystyle =\pi \Bigg[9\Bigg(\frac{2^3}{3} -\frac{1^3}{3} \Bigg)+54 \Bigg(\frac{2^2}{2} -\frac{1^2}{2} \Bigg)+81(2-1)\Bigg]=\\\\\\ =\pi \Bigg(9 \cdot \frac{8-1}{3} +54 \cdot \frac{4-1}{2} +81 \cdot 1\Bigg)=\pi\Bigg(\not9 \cdot \frac{7}{\not3} +\not54 \cdot \frac{3}{\not2} +81\Bigg)=\\\\\\=\pi\big(3 \cdot 7+27 \cdot 3+81\big)=\pi\big(21+81+81\big)=\underline{\underline{183 \pi}}$