Corrigé
1.
• f(x) = x5 – x4 + 2 donc F(x) = 1
6 x6 – 1
5 x5 + 2x
• g(x) = x6 – x3 + 2x donc G(x) = 1
7 x7 – 1
4 x4 + x2
• h(x) = 2
x2 – 4x3 + 3x donc H(x) = – 2
x – x4 + 3 2 x2
• k(x) = 3
x2 – 5x3 + 2x2 donc K(x) = – 3 x – 5
4 x4 + 2 3 x3
• l(x) = cos(5x – 3) donc L(x) = 1
5 sin(5x – 3)
• m(x) = sin(3x – 2 ) donc M(x) = – 1
3 cos(3x – 2)
• n(x) = 1
(3 x – 2)2 donc N(x) = – 1
3 × 1 3 x – 2
• p(x) = 2
(5 x – 3)2 donc P(x) = – 1
5 × 2 5 x – 3
• r(x) = 3
√
2 x – 7 donc R(x) = 3 × 2 × 12 ×
√
2 x – 7 = 3 ×√
2 x – 7• s(x) = 7
√
2 x+3 donc S(x) = – 7 × 2 × ( – 12 ) ×
√
2 x+3 = 7 ×√
2 x+32. Montrer que F(x) = (x3+7)2
2 est une primitive de f(x) = 3x2 (x3 + 7).
On dérive F.
F ' (x) = 1
2 × 2 (x3+7)1 × 3x2 = 3x2 (x3 + 7) = f(x).
F est bien une primitive de f.
