integral batas atas 2 bawah 1 (2x+3)²
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integral batas atas 2 bawah 1 (2x+3)² dx
= ∫(2x+3)(2x+3) dx
= ∫2x(2x+3) + 3(2x+3) dx
= ∫4x²+6x+6x+9 dx
= ∫4x²+12x+9 dx
= 4/2+1 x²+¹ + 12/1+1 x¹+¹ + 9x
= 4/3x³ + 12/2x² + 9x
= 4/3x³ + 6x² + 9x [2 1]
= (4/3(2)³ + 6(2)² + 9(2)) - (4/3(1)³ + 6(1)² + 9(1))
= (4/3(8) + 24 + 18) - (4/3 + 6 + 9)
= (32/3 + 42) - (4/3 + 15)
= 158/3 - 49/3
= 109/3
= 36 1/3
semoga membantu, semangat belajar!
