Question

tìm nguyên hàm F(x) của hàm số f(x) thỏa mãn điều kiện tương ứng

a) \(f\left(x\right)=x-cosx\) và \(F\left(\dfrac{\pi}{2}\right)=1\)

b) \(f\left(x\right)=2x-sinx\) và \(F\left(\dfrac{\pi}{4}\right)=1\)

\(F\left(\dfrac{\pi}{2}\right)=1\)

Answer 1

a: f(x)=x-cosx

=>\(F\left(x\right)=\int\left(x-cosx\right)dx=\dfrac{1}{2}x^2-sinx+C\)

\(F\left(\dfrac{\Omega}{2}\right)=1\)

=>\(\dfrac{1}{2}\cdot\left(\dfrac{\Omega}{2}\right)^2-sin\left(\dfrac{\Omega}{2}\right)+C=1\)

=>\(C=2-\dfrac{1}{2}\cdot\dfrac{\Omega^2}{4}=\dfrac{16-\Omega^2}{8}\)

=>\(F\left(x\right)=\dfrac{1}{2}x^2-sinx+\dfrac{16-\Omega^2}{8}\)

b: f(x)=2x-sinx

=>\(F\left(x\right)=\int\left(2x-sinx\right)dx=2\cdot\dfrac{1}{2}x^2+cosx+C=x^2+cosx+C\)

\(F\left(\dfrac{\Omega}{4}\right)=1\)

=>\(\left(\dfrac{\Omega}{4}\right)^2+cos\left(\dfrac{\Omega}{4}\right)+C=1\)

=>\(C=\dfrac{2-\sqrt{2}}{2}-\dfrac{\Omega^2}{16}=\dfrac{16-8\sqrt{2}-\Omega^2}{16}\)

=>\(F\left(x\right)=x^2+cosx+\dfrac{16-8\sqrt{2}-\Omega^2}{16}\)