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Câu hỏi: Nguyên hàm $F\left( x \right)$ của hàm số $f\left( x \right)=x+\sin x$ thỏa mãn $F\left( 0 \right)=19$ là:
A. $F\left( x \right)=-\cos x+\dfrac{{{x}^{2}}}{2}.$
B. $F\left( x \right)=-\cos x+\dfrac{{{x}^{2}}}{2}+20.$
C. $F\left( x \right)=-\cos x+\dfrac{{{x}^{2}}}{2}+2.$
D. $F\left( x \right)=\cos x+\dfrac{{{x}^{2}}}{2}+20.$
A. $F\left( x \right)=-\cos x+\dfrac{{{x}^{2}}}{2}.$
B. $F\left( x \right)=-\cos x+\dfrac{{{x}^{2}}}{2}+20.$
C. $F\left( x \right)=-\cos x+\dfrac{{{x}^{2}}}{2}+2.$
D. $F\left( x \right)=\cos x+\dfrac{{{x}^{2}}}{2}+20.$
$F\left( x \right)=\int\limits_{{}}^{{}}{f\left( x \right)dx}=\int\limits_{{}}^{{}}{\left( x+\sin x \right)dx}=\dfrac{{{x}^{2}}}{2}-\cos x+C.$
Theo bài $F\left( 0 \right)=19\Leftrightarrow \dfrac{{{0}^{2}}}{2}-\cos 0+C=19\Leftrightarrow C=20.$
Vậy $F\left( x \right)=\dfrac{{{x}^{2}}}{2}-\cos x+20.$
Theo bài $F\left( 0 \right)=19\Leftrightarrow \dfrac{{{0}^{2}}}{2}-\cos 0+C=19\Leftrightarrow C=20.$
Vậy $F\left( x \right)=\dfrac{{{x}^{2}}}{2}-\cos x+20.$
Đáp án B.