Jeg fikk som svar 5ln4+4, men det er feil....
svaret skal være 10ln2+4
Bestemt Integral
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Løser det ubestemt først.
[tex]\int (\frac{5}{x} + \frac{2}{\sqrt{x}})\rm{d}x = \int \frac{5}{x}\rm{d}x + \int \frac{2}{\sqrt{x}}\rm{d}x = 5\int \frac{1}{x}\rm{d}x + 2\int x^{-\frac{1}{2}}\rm{d}x[/tex]
[tex]5\int \frac{1}{x} \rm{d}x = 5\ln{x} + C[/tex]
[tex]2\int x^{-\frac{1}{2}}\rm{d}x = 2 \ \cdot \ 2\sqrt{x} + C[/tex]
[tex]\int (\frac{5}{x} + \frac{2}{\sqrt{x}})\rm{d}x = 5\ln{x} + 4\sqrt{x} + C[/tex]
Setter inn grenser:
[tex]\int_1^4 (\frac{5}{x} + \frac{2}{\sqrt{x}})\rm{d}x = \large\left[5\ln{x} + 4\sqrt{x} \large\right]_1^4 = 5\ln{4} + 8 - (5\ln{1} + 4) = 5\ln{2^2} + 8 - 4 = 5 \ \cdot \ 2 \ln{2} + 4 = \underline{\underline{10\ln{2} + 4}}[/tex]
[tex]\int (\frac{5}{x} + \frac{2}{\sqrt{x}})\rm{d}x = \int \frac{5}{x}\rm{d}x + \int \frac{2}{\sqrt{x}}\rm{d}x = 5\int \frac{1}{x}\rm{d}x + 2\int x^{-\frac{1}{2}}\rm{d}x[/tex]
[tex]5\int \frac{1}{x} \rm{d}x = 5\ln{x} + C[/tex]
[tex]2\int x^{-\frac{1}{2}}\rm{d}x = 2 \ \cdot \ 2\sqrt{x} + C[/tex]
[tex]\int (\frac{5}{x} + \frac{2}{\sqrt{x}})\rm{d}x = 5\ln{x} + 4\sqrt{x} + C[/tex]
Setter inn grenser:
[tex]\int_1^4 (\frac{5}{x} + \frac{2}{\sqrt{x}})\rm{d}x = \large\left[5\ln{x} + 4\sqrt{x} \large\right]_1^4 = 5\ln{4} + 8 - (5\ln{1} + 4) = 5\ln{2^2} + 8 - 4 = 5 \ \cdot \ 2 \ln{2} + 4 = \underline{\underline{10\ln{2} + 4}}[/tex]