Nguyên hàm \(F\left( x \right)\) của hàm số \(f\left( x \right) = 3{x^2} + 2x + 5\) thỏa mãn \(F\left( 1 \right) = 4\) là    

a) Đ, b) Đ, c) Đ, d) S

a) Có \(F'\left( x \right) = {\left( {\frac{{{x^3}}}{3}} \right)^\prime } = {x^2}\).

b) $\underset{0}{\overset{2}{\int }}\left(f\left(x\right)+1\right)dx=\underset{0}{\overset{2}{\int }}\left(x^2+1\right)dx={\left.\left(\frac{x^3}{3}+x\right)\right|}_0^2=\frac{14}{3}$

c) \(\int\limits_{2024}^a {{f^2}\left( x \right)dx} = \int\limits_{2024}^a {{x^4}dx} = \left. {\frac{{{x^5}}}{5}} \right|_{2024}^a = \frac{{{a^5}}}{5} - \frac{{{{2024}^5}}}{5} = 0 \Rightarrow a = 2024\).

Do đó \(2a - 1 = 4047\).

d) Ta có \(S = \int\limits_0^2 {\left| {{x^2}} \right|dx} = \int\limits_0^2 {{x^2}dx} = \left. {\frac{{{x^3}}}{3}} \right|_0^2 = \frac{8}{3}\).

Trả lời: 0,75

Ta có \(F\left( x \right) = \int {\sin 2xdx} = - \frac{1}{2}\cos 2x + C\).

Vì \(F\left( {\frac{\pi }{4}} \right) = 1\) nên \( - \frac{1}{2}\cos \frac{\pi }{2} + C = 1 \Rightarrow C = 1\).

Do đó \(F\left( x \right) = - \frac{1}{2}\cos 2x + 1\).

Vậy \(F\left( {\frac{\pi }{6}} \right) = - \frac{1}{2}\cos \frac{\pi }{3} + 1 = \frac{3}{4} = 0,75\).