Question

Bantu jawab pakai cara dan rumus​

Answer 1

Penjelasan dengan langkah-langkah:

[tex]5) \: \int _{2} {}^{4} \: x(3 + {x}^{3} {)}^{3} dx \\ = \int \: x \times (27 + 27 {x}^{3} + 9 {x}^{6} + {x}^{9} )dx \\ = \int27x + 27 {x}^{4} + 9 {x}^{7} + {x}^{10} )dx \\ = \int \: 27x \: dx + \int \: {27x}^{4} \: dx \: + \int \: 9 {x}^{7} \: dx \: + \int \: {x}^{10} \: dx \\ = \frac{27 {x}^{2} }{2} + \frac{27 {x}^{5} }{5} + \frac{ {9x}^{8} }{8} + \frac{ {x}^{11} }{11} \\ \\ = \frac{27 {x}^{2} }{2} + \frac{27 {x}^{5} }{5} + \frac{ {9x}^{8} }{8} + \frac{ {x}^{11} }{11} \: \: \huge{ | }_{2} {}^{4} \\ \\ = \frac{27 { \times 4}^{2} }{2} + \frac{27 { \times 4}^{5} }{5} + \frac{ {9 \times 4}^{8} }{8} + \frac{ {4}^{11} }{11} - \frac{27 { \times 2}^{2} }{2} + \frac{27 { \times 2}^{5} }{5} + \frac{ {9 \times 2}^{8} }{8} + \frac{ {2}^{11} }{11} \\ \\ = \frac{4332494}{55} + \frac{ {4}^{11} }{11} \\ = 460072.98[/tex]

6. Ada di foto

Answer 2

Penjelasan dengan langkah-langkah:

Nomor 5 :

[tex]\tt \int\limits^4_2 {x(3+x^3)^3} \, dx \\\\=\int\limits^4_2 {(27(x)+27x^{3+1}+9x^{6+1}+x^{9+1})} \, dx \\\\= \frac{27x^{1+1}}{1+1}+\frac{27x^{4+1}}{4+1}+\frac{9x^{7+1}}{7+1}+\frac{x^{10+1}}{10+1}\\ \\ = \frac{27x^2}{2}+\frac{27x^5}{5} +\frac{9x^8}{8}+\frac{x^{11}}{11}|^4_2\\[/tex]

[tex]= \frac{27(4)^2}{2}+\frac{27(4)^5}{5}+\frac{9(4)^8}{8}+\frac{4^{11}}{11}-(\frac{27(2)^2}{2}+ \frac{27(2)^5}{5}+\frac{9(2)^8}{8}+\frac{2^{11}}{11})\\ \\ = \frac{27(16)}{2}+\frac{27(1.024)}{5}+\frac{9(65.536)}{8}+\frac{4.194.304}{11}- (\frac{27(4)}{2}+\frac{27(32)}{5}+\frac{9(256)}{8}+\frac{2.048}{11})\\ \\ = \frac{4.332.494}{55}+\frac{4^{11}}{11}+C[/tex]

Nomor 6 :

[tex]\tt \int\limits {4x\sqrt{1+2x} } \, dx \\\\=4\int\limits{x\sqrt{1+2x} } \, dx \\\\=4\int\limits {x\sqrt{1+2x} \times \frac{1}{2} }\, dt \\\\=4\int\limits{\frac{\frac{(1+2x)-1}{2}\times \sqrt{1+2x} }{2}} \, dt \\\\= 4\int\limits {\frac{(1+2x)-1\sqrt{1+2x}}{4}} \, dt \\[/tex]

[tex]\tt =\int\limits(1+2x)^{1+\frac{1}{2}}-(1+2x)^\frac{1}{2} \, dt \\\\= \frac{(1+2x)^{\frac{3}{2}+1}}{\frac{3}{2}+1}-\frac{(1+2x)^{\frac{1}{2} +1}}{\frac{1}{2}+1}\\ \\ =\frac{2(1+2x)^\frac{5}{2}}{5}-\frac{2(1+2x)^\frac{3}{2}}{3}\\ \\ =\frac{2\sqrt{1+2x}(1+2x)^2}{5}-\frac{2(1+2x)\sqrt{1+2x}}{3}\\ \\ =\frac{2\sqrt{1+2x}(1+4x+4x^2)}{5}- \frac{2(1+2x)\sqrt{1+2x}}{3}+C[/tex]