Luas daerah yang dibatasi garis y = x + 3, sumbu X, x=1, dan x=a (a>1) adalah 10 satuan luas. nilai a adalah ....
Takamori37
Artinya hanyalah,
![\int_1^a x+3$ dx$ =10 \\ \frac{1}{2}x^2+3x]_1^a =10 \\ (\frac{1}{2}a^2+3a)-(\frac{1}{2}1^2+3.1) =10 \\ \frac{1}{2}a^2+3a-\frac{7}{2}=10 \\ a^2+6a-7=20 \\ a^2+6a-27=0 \\ (a+9)(a-3)=0 \\ a=-9$ (Tidak mungkin, karena diminta $ a>0 \\ a=3](https://tex.z-dn.net/?f=%5Cint_1%5Ea+x%2B3%24+dx%24+%3D10+%5C%5C+%5Cfrac%7B1%7D%7B2%7Dx%5E2%2B3x%5D_1%5Ea+%3D10+%5C%5C+%28%5Cfrac%7B1%7D%7B2%7Da%5E2%2B3a%29-%28%5Cfrac%7B1%7D%7B2%7D1%5E2%2B3.1%29+%3D10+%5C%5C+%5Cfrac%7B1%7D%7B2%7Da%5E2%2B3a-%5Cfrac%7B7%7D%7B2%7D%3D10+%5C%5C+a%5E2%2B6a-7%3D20+%5C%5C+a%5E2%2B6a-27%3D0+%5C%5C+%28a%2B9%29%28a-3%29%3D0+%5C%5C+a%3D-9%24+%28Tidak+mungkin%2C+karena+diminta+%24+a%3E0+%5C%5C+a%3D3 "\int_1^a x+3$ dx$ =10 \\ \frac{1}{2}x^2+3x]_1^a =10 \\ (\frac{1}{2}a^2+3a)-(\frac{1}{2}1^2+3.1) =10 \\ \frac{1}{2}a^2+3a-\frac{7}{2}=10 \\ a^2+6a-7=20 \\ a^2+6a-27=0 \\ (a+9)(a-3)=0 \\ a=-9$ (Tidak mungkin, karena diminta $ a>0 \\ a=3")
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