Penjelasan dengan langkah-langkah:

INTEGRAL TAK TENTU DAN INTEGRAL TENTU

dy/dx = 4x -6

\int( \frac{dy}{dx}) dx = \int (4x^1 -6x^0) dx∫(

dx

dy

)dx=∫(4x

1

−6x

0

)dx

y = \frac{4}{1 + 1} x^{1 + 1} -\frac{6}{0 + 1}x^{0 + 1} + C

1+1

4

x

1+1

−

0+1

6

x

0+1

+C

y = f(x) = 2x² -6x + C

melalui P (2, 1) maka f(2) = 1

2(2)² -6(2) + C = 1

8 -12 + C = 1

C -4 = 1

C = 5

f(x) = 2x² -6x + C

f(x) = y = 2x² -6x + 5

[Opsi A]

\displaystyle \int^2_0 (2x + 1)(3x -1) \: dx∫

0

2

(2x+1)(3x−1)dx

= \displaystyle \int^2_0 (6x^2 + x -1) \: dx∫

0

2

(6x

2

+x−1)dx

= [ \frac{6}{2 + 1}x^{2 + 1} + \frac{1}{1 + 1}x^{1 + 1} - \frac{1}{0 + 1}x^{0 + 1} ]^2_0[

2+1

6

x

2+1

+

1+1

1

x

1+1

−

0+1

1

x

0+1

]

0

2

= [2x^3 + \frac{1}{2} x^2 -x]^2_0[2x

3

+

2

1

x

2

−x]

0

2

= (2(2)³ + ½(2)² -2) -(2(0)³ + ½(0)² -0)

= (16 + 2 -2) -(0)

= 16

[Opsi A]