Diketahui F"(x)=12x^2+6x-10

Diketahui $F^{″} (x) = 12 x^{2} + 6 x - 10$ . Jika $F^{'} (- 1) = 5$ dan $F (2) = 3, F (x) = \dots$

A. $4 x^{3} + 3 x^{2} - 10 x - 4$

B. $4 x^{3} + 3 x^{2} - 10 x + 4$

C. $x^{4} + x^{3} - 5 x^{2} - 4 x + 7$

D. $x^{4} + x^{3} - 5 x^{2} + 7 x - 4$

E. $x^{4} + x^{3} - 5 x^{2} + 4 x - 7$

Penyelesaian

KONSEP

$\int f^{″} (x) d x = f^{'} (x)$

$\int f^{'} (x) d x = f (x)$

Jawab

Mencari $F^{'} (X)$

$\int (12 x^{2} + 6 x - 10) d x$

$F^{'} (x) = \frac{12}{3} x^{3} + \frac{6}{2} x^{2} + 10 x + C$

$F^{'} (x) = 4 x^{3} + 3 x^{2} - 10 x + C$

$F^{'} (- 1) = 4 (- 1)^{3} + 3 (- 1)^{2} - 10 (- 1) + C$

$5 = - 4 + 3 + 10 + C$

$5 = 9 + C$

$- 4 = C$

$f^{'} (x) = 4 x^{3} + 3 x^{2} - 10 x - 4$

Mencari $F (x)$

$\int (4 x^{3} + 3 x^{2} - 10 x - 4) d x$

$F (x) = \frac{4}{4} x^{4} + \frac{3}{3} x^{3} - \frac{10}{2} x^{2} - 4 x + C$

$F (x) = x^{4} + x^{3} - 5 x^{2} - 4 x + C$

$F (2) = 3$

$x^{4} + x^{3} - 5 x^{2} - 4 x + C = 3$

$(2)^{4} + (2)^{3} - 5 (2)^{2} - 4 (2) + C = 3$

$16 + 8 - 20 - 8 + C = 3$

$- 4 + C = 3$

$C = 7$

$F (x) = x^{4} + x^{3} - 5 x^{2} - 4 x + 7$

Jawaban: C

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