Упр.23.3 ГДЗ Мордкович Семенов 11 класс (Алгебра)

а) рис. 113, а; б) рис. 113, б. Найдите площадь криволинейной трапеции, ограниченной прямыми y=0, x=a, x=b и графиком функции y=(x):
а) y=cos(x), a=0, b=?/2; б) y=sin(2x), a=?/6, b=?/3;
в) y=1+2cos(2x), a=?/12, b=?/4; г) y=sin(x), a=?/4, b=3?/4;
д) y=cos(x/2), a=?/2, b=?; е) y=1-(1/2)sin(x/2), a=2?/3, b=?.

1. а)

   $$S=\int_{0}^{\pi/2}\cos x\,dx=\sin x\Big|_{0}^{\pi/2}$$
   $$S=\sin\frac{\pi}{2}-\sin 0=1.$$

2. б)

   $$S=\int_{\pi/6}^{\pi/3}\sin 2x\,dx=-\frac12\cos 2x\Big|_{\pi/6}^{\pi/3}$$
   $$S=-\frac12\cos\frac{2\pi}{3}+\frac12\cos\frac{\pi}{3}$$
   $$S=-\frac12\left(-\frac12\right)+\frac12\cdot\frac12=\frac12.$$

3. в)

   $$S=\int_{\pi/12}^{\pi/4}(1+2\cos 2x)\,dx=(x+\sin 2x)\Big|_{\pi/12}^{\pi/4}$$
   $$S=\left(\frac{\pi}{4}+\sin\frac{\pi}{2}\right)-\left(\frac{\pi}{12}+\sin\frac{\pi}{6}\right)$$
   $$S=\frac{\pi}{4}+1-\frac{\pi}{12}-\frac12=\frac12+\frac{\pi}{6}.$$

4. г)

   $$S=\int_{\pi/4}^{3\pi/4}\sin x\,dx=-\cos x\Big|_{\pi/4}^{3\pi/4}$$
   $$S=-\cos\frac{3\pi}{4}+\cos\frac{\pi}{4}$$
   $$S=-\left(-\frac{\sqrt2}{2}\right)+\frac{\sqrt2}{2}=\sqrt2.$$

5. д)

   $$S=\int_{\pi/2}^{\pi}\cos\frac{x}{2}\,dx=2\sin\frac{x}{2}\Big|_{\pi/2}^{\pi}$$
   $$S=2\sin\frac{\pi}{2}-2\sin\frac{\pi}{4}$$
   $$S=2-\sqrt2.$$

6. е)

   $$S=\int_{2\pi/3}^{\pi}\left(1-\frac12\sin\frac{x}{2}\right)\,dx=\left(x+\cos\frac{x}{2}\right)\Big|_{2\pi/3}^{\pi}$$
   $$S=\left(\pi+\cos\frac{\pi}{2}\right)-\left(\frac{2\pi}{3}+\cos\frac{\pi}{3}\right)$$
   $$S=\pi-\frac{2\pi}{3}-\frac12=\frac{\pi}{3}-\frac12.$$

Ответ

а) $$1$$; б) $$\frac12$$; в) $$\frac12+\frac{\pi}{6}$$; г) $$\sqrt2$$; д) $$2-\sqrt2$$; е) $$\frac{\pi}{3}-\frac12$$.