Page 54 - E-Book Kalkulus Integral

Langkah 4: Menyelesaikan perhitungan Luas menggunakan Integral Tentu

−5
( ) = ∫ −5,18 (16 + 83)
1
2
= [8 + 83 ] −5
−5,18
= (−215) − (−215,2808) = 0,2808

5 1
( ) = ∫ −5 10 ( + 35)
2
1
5
2
= [ + 70 ]
20 −5
1
= ((375) − (−325)) = 35
20

5 1 −5,18
( ) = (− ∫ (4 − 83) ) − (− ∫ (16 + 83) )
3
−5,5 21 −5,5
1
2
2
= − (2 − 83 ) 5 + [8 + 83 ] −5,18
21 −5,5 −5,5
1
= − (−365 − 571) + (−215,2808 − (−214,5))
21
= 42 + (−0,7808) = 41,2192

Sehingga

( ) = ( ) + ( ) + ( ) = 0,2808 + 35 + 41,2192 = 76,5
1
3
2
Kesimpulan
Karena perbandingan gambar dengan ukuran asli 7:50 maka

7 2 76,5
( ) =
50 ( )
49 = 76,5
2500 ( )

2500 .76,5 2
( ) = = 3903,0612
49

Kalkulus Integral berbasis Project Based Learning 50