Luas daerah yang dibatasi oleh kurva f(x) = -x^2+8x-15, sumbu x, dan sumbu y adalah
A. 19
Satuan Luas
B.21
SL
C.24
SL
D.25 SL
E. 28 SL
A. 19
B.21
C.24
D.25
E. 28
izharulhaq78
![\displaystyle L_1=\int\limits^5_3f(x)\,dx\\L_1=\int\limits^5_3-x^2+8x-15\,dx\\L_1=\left-\frac{1}{2+1}x^{2+1}+\frac{8}{1+1}x^{1+1}-\frac{15}{0+1}x^{0+1}\right|^5_3\\L_1=\left-\frac{1}{3}x^{3}+\frac{8}{2}x^{2}-\frac{15}{1}x^{1}\right|^5_3\\L_1=-\frac{1}{3}(5)^{3}+4(5)^{2}-15(5)-\left[-\frac{1}{3}(3)^{3}+4(3)^{2}-15(3)\right]\\L_1=-\frac{125}{3}+100-75-[-9+36-45]\\L_1=-\frac{125}{3}+25-[-18]\\L_1=-\frac{125}{3}+\frac{3\times43}{3}\\L_1=-\frac{125}{3}+\frac{129}{3}\\L_1=\frac{4}{3}\\L_1=1\frac{1}{3}](https://tex.z-dn.net/?f=%5Cdisplaystyle+L_1%3D%5Cint%5Climits%5E5_3f%28x%29%5C%2Cdx%5C%5CL_1%3D%5Cint%5Climits%5E5_3-x%5E2%2B8x-15%5C%2Cdx%5C%5CL_1%3D%5Cleft-%5Cfrac%7B1%7D%7B2%2B1%7Dx%5E%7B2%2B1%7D%2B%5Cfrac%7B8%7D%7B1%2B1%7Dx%5E%7B1%2B1%7D-%5Cfrac%7B15%7D%7B0%2B1%7Dx%5E%7B0%2B1%7D%5Cright%7C%5E5_3%5C%5CL_1%3D%5Cleft-%5Cfrac%7B1%7D%7B3%7Dx%5E%7B3%7D%2B%5Cfrac%7B8%7D%7B2%7Dx%5E%7B2%7D-%5Cfrac%7B15%7D%7B1%7Dx%5E%7B1%7D%5Cright%7C%5E5_3%5C%5CL_1%3D-%5Cfrac%7B1%7D%7B3%7D%285%29%5E%7B3%7D%2B4%285%29%5E%7B2%7D-15%285%29-%5Cleft%5B-%5Cfrac%7B1%7D%7B3%7D%283%29%5E%7B3%7D%2B4%283%29%5E%7B2%7D-15%283%29%5Cright%5D%5C%5CL_1%3D-%5Cfrac%7B125%7D%7B3%7D%2B100-75-%5B-9%2B36-45%5D%5C%5CL_1%3D-%5Cfrac%7B125%7D%7B3%7D%2B25-%5B-18%5D%5C%5CL_1%3D-%5Cfrac%7B125%7D%7B3%7D%2B%5Cfrac%7B3%5Ctimes43%7D%7B3%7D%5C%5CL_1%3D-%5Cfrac%7B125%7D%7B3%7D%2B%5Cfrac%7B129%7D%7B3%7D%5C%5CL_1%3D%5Cfrac%7B4%7D%7B3%7D%5C%5CL_1%3D1%5Cfrac%7B1%7D%7B3%7D "\displaystyle L_1=\int\limits^5_3f(x)\,dx\\L_1=\int\limits^5_3-x^2+8x-15\,dx\\L_1=\left-\frac{1}{2+1}x^{2+1}+\frac{8}{1+1}x^{1+1}-\frac{15}{0+1}x^{0+1}\right|^5_3\\L_1=\left-\frac{1}{3}x^{3}+\frac{8}{2}x^{2}-\frac{15}{1}x^{1}\right|^5_3\\L_1=-\frac{1}{3}(5)^{3}+4(5)^{2}-15(5)-\left[-\frac{1}{3}(3)^{3}+4(3)^{2}-15(3)\right]\\L_1=-\frac{125}{3}+100-75-[-9+36-45]\\L_1=-\frac{125}{3}+25-[-18]\\L_1=-\frac{125}{3}+\frac{3\times43}{3}\\L_1=-\frac{125}{3}+\frac{129}{3}\\L_1=\frac{4}{3}\\L_1=1\frac{1}{3}")
![\displaystyle L_2=-\int\limits^3_0f(x)\,dx\\L_2=-\int\limits^3_0-x^2+8x-15\,dx\\L_2=-\left(\left-\frac{1}{2+1}x^{2+1}+\frac{8}{1+1}x^{1+1}-\frac{15}{0+1}x^{0+1}\right|^3_0\right)\\L_2=-\left(\left-\frac{1}{3}x^{3}+\frac{8}{2}x^{2}-\frac{15}{1}x^{1}\right|^3_0\right)\\L_2=-\left(-\frac{1}{3}(3)^{3}+4(3)^{2}-15(3)-\left[-\frac{1}{3}(0)^{3}+4(0)^{2}-15(0)\right]\right)\\L_2=-(-9+36-45-[0])\\L_2=-(-18)\\L_2=18](https://tex.z-dn.net/?f=%5Cdisplaystyle+L_2%3D-%5Cint%5Climits%5E3_0f%28x%29%5C%2Cdx%5C%5CL_2%3D-%5Cint%5Climits%5E3_0-x%5E2%2B8x-15%5C%2Cdx%5C%5CL_2%3D-%5Cleft%28%5Cleft-%5Cfrac%7B1%7D%7B2%2B1%7Dx%5E%7B2%2B1%7D%2B%5Cfrac%7B8%7D%7B1%2B1%7Dx%5E%7B1%2B1%7D-%5Cfrac%7B15%7D%7B0%2B1%7Dx%5E%7B0%2B1%7D%5Cright%7C%5E3_0%5Cright%29%5C%5CL_2%3D-%5Cleft%28%5Cleft-%5Cfrac%7B1%7D%7B3%7Dx%5E%7B3%7D%2B%5Cfrac%7B8%7D%7B2%7Dx%5E%7B2%7D-%5Cfrac%7B15%7D%7B1%7Dx%5E%7B1%7D%5Cright%7C%5E3_0%5Cright%29%5C%5CL_2%3D-%5Cleft%28-%5Cfrac%7B1%7D%7B3%7D%283%29%5E%7B3%7D%2B4%283%29%5E%7B2%7D-15%283%29-%5Cleft%5B-%5Cfrac%7B1%7D%7B3%7D%280%29%5E%7B3%7D%2B4%280%29%5E%7B2%7D-15%280%29%5Cright%5D%5Cright%29%5C%5CL_2%3D-%28-9%2B36-45-%5B0%5D%29%5C%5CL_2%3D-%28-18%29%5C%5CL_2%3D18 "\displaystyle L_2=-\int\limits^3_0f(x)\,dx\\L_2=-\int\limits^3_0-x^2+8x-15\,dx\\L_2=-\left(\left-\frac{1}{2+1}x^{2+1}+\frac{8}{1+1}x^{1+1}-\frac{15}{0+1}x^{0+1}\right|^3_0\right)\\L_2=-\left(\left-\frac{1}{3}x^{3}+\frac{8}{2}x^{2}-\frac{15}{1}x^{1}\right|^3_0\right)\\L_2=-\left(-\frac{1}{3}(3)^{3}+4(3)^{2}-15(3)-\left[-\frac{1}{3}(0)^{3}+4(0)^{2}-15(0)\right]\right)\\L_2=-(-9+36-45-[0])\\L_2=-(-18)\\L_2=18")
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