Jawaban:
integral tentu
[tex] \int \: ax {}^{n} \: dx \to \: \frac{ {ax}^{n + 1} }{n + 1} + C \\ [/tex]
.
[tex] \int_{2}^{5} 3 {x}^{2} + 4 \: dx \\ [/tex]
[tex] \int \: 3 {x }^{2} + 4 \: dx \\ \frac{3 {x}^{2 + 1} }{2 + 1} + 4x + C \\ \frac{3}{3} {x}^{3} + 4x + C = {x}^{3} + 4x + C[/tex]
batas atas= 5 dan batas bawah = 2
maka:
[tex]( {x}^{3} + 4x) - ( {x}^{3} + 4x) = ( {5}^{3} + 4(5)) - ( {2}^{3} + 4(2)) = (125 + 20 - 8 - 8) = 145 - 16 = 129[/tex]
[tex]\displaystyle\int_2^5\left(3x^2+4\right)dx=\boxed{\bf129}[/tex]
Penjelasan dengan langkah-langkah
Integral Tentu
[tex]\begin{aligned}\int_2^5\left(3x^2+4\right)dx&=\left[\frac{3x^{2+1}}{2+1}+4x\right]_2^5\\&=\left[\frac{3x^{2+1}}{3}+4x\right]_2^5\\&=\Bigl[x^3+4x\Bigr]_2^5\\&=\left(5^3+4\cdot5\right)-\left(2^3+4\cdot2\right)\\&=(125+20)-(8+8)\\&=145-16\\&=\bf129\end{aligned}[/tex]
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Jawaban:
integral tentu
[tex] \int \: ax {}^{n} \: dx \to \: \frac{ {ax}^{n + 1} }{n + 1} + C \\ [/tex]
.
[tex] \int_{2}^{5} 3 {x}^{2} + 4 \: dx \\ [/tex]
[tex] \int \: 3 {x }^{2} + 4 \: dx \\ \frac{3 {x}^{2 + 1} }{2 + 1} + 4x + C \\ \frac{3}{3} {x}^{3} + 4x + C = {x}^{3} + 4x + C[/tex]
batas atas= 5 dan batas bawah = 2
maka:
[tex]( {x}^{3} + 4x) - ( {x}^{3} + 4x) = ( {5}^{3} + 4(5)) - ( {2}^{3} + 4(2)) = (125 + 20 - 8 - 8) = 145 - 16 = 129[/tex]
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[tex]\displaystyle\int_2^5\left(3x^2+4\right)dx=\boxed{\bf129}[/tex]
Penjelasan dengan langkah-langkah
Integral Tentu
[tex]\begin{aligned}\int_2^5\left(3x^2+4\right)dx&=\left[\frac{3x^{2+1}}{2+1}+4x\right]_2^5\\&=\left[\frac{3x^{2+1}}{3}+4x\right]_2^5\\&=\Bigl[x^3+4x\Bigr]_2^5\\&=\left(5^3+4\cdot5\right)-\left(2^3+4\cdot2\right)\\&=(125+20)-(8+8)\\&=145-16\\&=\bf129\end{aligned}[/tex]