Упр.393 ГДЗ Колягин Ткачёва 11 класс (Алгебра)

1. $$\int_0^1 (5x^4-8x^3)\,dx$$
2. $$\int_{-1}^1 (6x^3-5x)\,dx$$
3. $$\int_1^4 \sqrt{x}\left(3-\frac{7}{x}\right)\,dx$$
4. $$\int_1^8 4\sqrt[3]{x}\left(1-\frac{4}{x}\right)\,dx$$
5. $$\int_0^3 \sqrt{x+1}\,dx$$
6. $$\int_2^6 \sqrt{2x-3}\,dx$$

1. $$\int_0^1 (5x^4-8x^3)\,dx=\left(x^5-2x^4\right)\Big|_0^1=(1-2)-(0-0)=-1.$$

2. $$\int_{-1}^1 (6x^3-5x)\,dx=\left(\frac{3}{2}x^4-\frac{5}{2}x^2\right)\Big|_{-1}^1.$$

   Подынтегральная функция нечётная, а пределы симметричны относительно нуля, значит
   $$\int_{-1}^1 (6x^3-5x)\,dx=0.$$

3. $$\int_1^4 \sqrt{x}\left(3-\frac{7}{x}\right)\,dx=\int_1^4 \left(3x^{1/2}-7x^{-1/2}\right)\,dx.$$

   $$=\left(2x^{3/2}-14x^{1/2}\right)\Big|_1^4=(16-28)-(2-14)=0.$$

4. $$\int_1^8 4\sqrt[3]{x}\left(1-\frac{4}{x}\right)\,dx=\int_1^8 \left(4x^{1/3}-16x^{-2/3}\right)\,dx.$$

   $$=\left(3x^{4/3}-48x^{1/3}\right)\Big|_1^8=(48-96)-(3-48)=-3.$$

5. $$\int_0^3 \sqrt{x+1}\,dx=\int_0^3 (x+1)^{1/2}\,dx.$$

   $$=\frac{2}{3}(x+1)^{3/2}\Big|_0^3=\frac{2}{3}\left(4^{3/2}-1^{3/2}\right)=\frac{2}{3}(8-1)=\frac{14}{3}.$$

6. $$\int_2^6 \sqrt{2x-3}\,dx.$$

   Сделаем замену: $$u=2x-3,\quad du=2\,dx,\quad dx=\frac{du}{2}.$$

   При $$x=2$$ имеем $$u=1$$, при $$x=6$$ имеем $$u=9$$. Тогда
   $$\int_2^6 \sqrt{2x-3}\,dx=\frac12\int_1^9 u^{1/2}\,du=\frac12\cdot\frac{2}{3}u^{3/2}\Big|_1^9.$$

   $$=\frac{1}{3}(9^{3/2}-1^{3/2})=\frac{1}{3}(27-1)=\frac{26}{3}.$$

Ответ

1) $$-1$$; 2) $$0$$; 3) $$0$$; 4) $$-3$$; 5) $$\frac{14}{3}$$; 6) $$\frac{26}{3}$$.