Упр.1036 ГДЗ Алимов 10-11 класс (Алгебра)
- 1036. Вычислить интеграл:
1) $$\int_{0}^{1}(5x^4-8x^3)\,dx$$;
2) $$\int_{-1}^{\frac{1}{2}}(6x^3-5x)\,dx$$;
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) $$\int_0^1 (5x^4-8x^3)\,dx=\left(x^5-2x^4\right)\Big|_0^1=1-2=-1.$$
2) $$\int_{-1}^2 (6x^3-5x)\,dx=\left(\frac{3x^4}{2}-\frac{5x^2}{2}\right)\Big|_{-1}^2=24-10-\frac{3}{2}+\frac{5}{2}=15.$$
3) $$\int_1^4 \sqrt{x}\left(3-\frac{7}{x}\right)\,dx=\int_1^4 \left(3x^{1/2}-7x^{-1/2}\right)\,dx$$
$$=\left(2x\sqrt{x}-14\sqrt{x}\right)\Big|_1^4=(16-28)-(2-14)=0.$$
4) $$\int_1^8 4\sqrt[3]{x}\left(1-\frac{4}{x}\right)\,dx=\int_1^8 \left(4x^{1/3}-16x^{-2/3}\right)\,dx$$
$$=\left(3x\sqrt[3]{x}-48\sqrt[3]{x}\right)\Big|_1^8=(48-96)-(3-48)=-3.$$
5) $$\int_0^3 \sqrt{x+1}\,dx=\int_0^3 (x+1)^{1/2}\,dx=\frac{2}{3}(x+1)^{3/2}\Big|_0^3$$
$$=\frac{2}{3}\left(4^{3/2}-1^{3/2}\right)=\frac{2}{3}(8-1)=\frac{14}{3}=4\frac{2}{3}.$$
6) $$\int_2^6 \sqrt{2x-3}\,dx=\frac{1}{2}\int_2^6 (2x-3)^{1/2}\,dx=\frac{1}{3}(2x-3)^{3/2}\Big|_2^6$$
$$=\frac{1}{3}\left(9^{3/2}-1^{3/2}\right)=\frac{1}{3}(27-1)=\frac{26}{3}=8\frac{2}{3}.$$
Ответ: 1) $$-1$$; 2) $$15$$; 3) $$0$$; 4) $$-3$$; 5) $$\frac{14}{3}$$; 6) $$\frac{26}{3}$$.
