Упр.1135 ГДЗ Колягин Ткачёва 11 класс (Алгебра)

1135. Вычислить интегралы:

1. $$\int_{2}^{9}\sqrt[3]{x-1}\,dx$$;
2. $$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}(2\cos 2x-1)\,dx$$;
3. $$\int_{3}^{4}(x^2+3/x-2)\,dx$$;
4. $$\int_{0}^{6}x\sqrt{36-x^2}\,dx$$.

1. $$\int\limits_{2}^{9}\sqrt[3]{x-1}\,dx=\int\limits_{2}^{9}(x-1)^{1/3}\,dx$$
   $$=\left(\frac{(x-1)^{4/3}}{4/3}\right)\Bigg|_{2}^{9} =\frac{3}{4}(x-1)^{4/3}\Bigg|_{2}^{9}$$
   $$=\frac{3}{4}\left(8^{4/3}-1^{4/3}\right) =\frac{3}{4}(16-1)=\frac{45}{4}=11{,}25.$$

2. $$\int\limits_{\pi/6}^{\pi/4}(2\cos^2 x-1)\,dx =\int\limits_{\pi/6}^{\pi/4}\cos 2x\,dx$$
   $$=\frac12\sin 2x\Bigg|_{\pi/6}^{\pi/4} =\frac12\left(\sin\frac{\pi}{2}-\sin\frac{\pi}{3}\right)$$
   $$=\frac12\left(1-\frac{\sqrt3}{2}\right) =\frac{2-\sqrt3}{4}.$$

3. $$\int\limits_{3}^{4}\frac{x^2+3}{x-2}\,dx =\int\limits_{3}^{4}\frac{x^2-4+7}{x-2}\,dx$$
   $$=\int\limits_{3}^{4}\left(x+2+\frac{7}{x-2}\right)\,dx$$
   $$=\left(\frac{x^2}{2}+2x+7\ln|x-2|\right)\Bigg|_{3}^{4}$$
   $$=\left(8+8+7\ln 2\right)-\left(\frac92+6+7\ln 1\right)$$
   $$=\frac72+7\ln 2.$$

4. $$\int\limits_{0}^{6}x\sqrt{36-x^2}\,dx$$
   Сделаем замену:
   $$u=36-x^2,\qquad du=-2x\,dx,\qquad x\,dx=-\frac12\,du.$$
   Тогда
   $$\int\limits_{0}^{6}x\sqrt{36-x^2}\,dx =-\frac12\int\limits_{0}^{6}\sqrt{u}\,du$$
   $$=-\frac12\cdot\frac{2}{3}u^{3/2}\Bigg|_{0}^{6} =-\frac13(36-x^2)^{3/2}\Bigg|_{0}^{6}$$
   $$=-\frac13\cdot 0+\frac13\cdot 36^{3/2} =\frac13\cdot 216=72.$$

Ответ

1) $$11{,}25$$; 2) $$\frac{2-\sqrt3}{4}$$; 3) $$\frac72+7\ln 2$$; 4) $$72$$.