HITUNGLAH NILAI TIAP INTEGRAL TERTENTU DI BAWAH INI 1) {3 bawahnya 1, ( 1 + 2x ) dx 2) { 5dibawahnya.... Question from @wennytaAyu

bawahnya semua itu bkn dibawahnya

Jadi misalkan atas 5 bawah 0 gitu?

iya benar bgt izhar:)

kok lama izhar

msh lama ya?????

2 no. lagi

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Jawabanmu

$1.$
$\int\limit^3_11+2x\,dx\\\frac{1}{0+1}x^{0+1}+\frac{2}{1+1}x^{1+1}]^3_1\\\frac{1}{1}x^{1}+\frac{2}{2}x^{2}]^3_1\\x+x^{2}]^3_1\\(3+3^{2})-(1+1^{2})\\(3+9)-(1+1)\\12-2\\\boxed{10}$

$2.$
$\int\limits^5_01+2x^3\,dx\\\frac{1}{0+1}x^{0+1}+\frac{2}{3+1}x^{3+1}]^5_0\\\frac{1}{1}x^{1}+\frac{2}{4}x^{4}]^5_0\\x+\frac{1}{2}x^{4}]^5_0\\5+\frac{1}{2}5^{4}-(0+\frac{1}{2}0^{4})\\5+\frac{625}{2}-(0+0)\\5+312,5\\\boxed{317,5}$

$3.$
$\int\limits^0_{-3}x^2+2x\,dx\\\frac{1}{2+1}x^{2+1}+\frac{2}{1+1}x^{1+1}]^0_{-3}\\\frac{1}{3}x^{3}+\frac{2}{2}x^{2}]^0_{-3}\\\frac{1}{3}0^{3}+0^{2}-(\frac{1}{3}(-3)^{3}+(-3)^{2})\\0+0-(\frac{1}{3}(-27)+9)\\0-(-9+9)\\0-0\\\boxed{0}$

$4.$
$\int\limits^5_3\sqrt{x}\,dx\\\int\limits^5_3x^{\frac{3}{2}}\,dx\\\frac{1}{\frac{3}{2}+1}x^{\frac{3}{2}+1}]^5_3\\\frac{1}{\frac{5}{2}}x^{\frac{5}{2}}]^5_3\\\frac{2}{5}x^{2}\sqrt{x}]^5_3\\\frac{2}{5}5^{2}\sqrt{5}-(\frac{2}{5}3^{2}\sqrt{3})\\\frac{2}{5}25\sqrt{5}-(\frac{2}{5}9\sqrt{3})\\\boxed{10\sqrt{5}-\frac{18}{5}\sqrt{3}}$

$5.$
$\int\limits^4_3\frac{1}{\sqrt{x}}\,dx\\\int\limits^4_3x^{-\frac{1}{2}}\,dx\\\frac{1}{-\frac{1}{2}+1}x^{-\frac{1}{2}+1}]^4_3\\\frac{1}{\frac{1}{2}}x^{\frac{1}{2}}]^4_3\\2\sqrt{x}]^4_3\\2\sqrt{4}-(2\sqrt{3})\\2\times2-2\sqrt{3}\\\boxed{4-2\sqrt{3}}$

$6.$
$\int\limits^3_0(x-1)^2\,dx\\\int\limits^3_0x^2-2x+1\,dx\\\frac{1}{2+1}x^{2+1}-\frac{2}{1+1}x^{1+1}+\frac{1}{0+1}x^{0+1}]^3_0\\\frac{1}{3}x^{3}-\frac{2}{2}x^{2}+\frac{1}{1}x^{1}]^3_0\\\frac{1}{3}3^{3}-3^{2}+3\\\frac{1}{3}(27)-9+3\\9-9+3\\\boxed{3}$

$7.$
$\int\limits^1_{-2}(x+2)^2\,dx\\\int\limits^1_{-2}x^2+4x+4\,dx\\\frac{1}{2+1}x^{2+1}+\frac{4}{1+1}x^{1+1}+\frac{4}{0+1}x^{0+1}]^1_{-2}\\\frac{1}{3}x^{3}+\frac{4}{2}x^{2}+\frac{4}{1}x^{1}]^1_{-2}\\\frac{1}{3}(1)^{3}+2(1)^{2}+4(1)-(\frac{1}{3}(-2)^{3}+2(-2)^{2}+4(-2))\\\frac{1}{3}+2+4-(\frac{1}{3}(-8)+2(4)-8)\\\frac{1}{3}+6+\frac{1}{3}(8)\\6+\frac{1}{3}(9)\\6+3\\\boxed{9}$

$8.$
$\int\limits^2_{-1}(x+1)(x+3)\,dx\\\int\limits^2_{-1}x^2+4x+3\,dx\\\frac{1}{2+1}x^{2+1}+\frac{4}{1+1}x^{1+1}+\frac{3}{0+1}x^{0+1}]^2_{-1}\\\frac{1}{3}x^{3}+2x^{2}+3x]^2_{-1}\\\frac{1}{3}(2)^{3}+2(2)^{2}+3(2)-(\frac{1}{3}(-1)^{3}+2(-1)^{2}+3(-1))\\\frac{1}{3}(8)+8+6-(-\frac{1}{3}+2-3)\\\frac{1}{3}(8)+14+\frac{1}{3}+1\\\frac{1}{3}(9)+15\\3+15\\\boxed{18}$

$9.$
$\int\limits^2_1x(x-1)(x-2)\,dx\\\int\limits^2_1x(x^2-3x+2)\,dx\\\int\limits^2_1x^3-3x^2+2x\,dx\\\frac{1}{3+1}x^{3+1}-\frac{3}{2+1}x^{2+1}+\frac{2}{1+1}x^{1+1}]^2_1\\\frac{1}{4}x^{4}-\frac{3}{3}x^{3}+\frac{2}{2}x^{2}]^2_1\\\frac{1}{4}2^{4}-2^{3}+2^{2}-(\frac{1}{4}1^{4}-1^{3}+1^{2})\\4-8+4-\frac{1}{4}\\0-\frac{1}{4}\\\boxed{-\frac{1}{4}}$

$10.$
$\int\limits^2_1x+\frac{1}{x^2}\,dx\\\int\limits^2_1x+x^{-2}\,dx\\\frac{1}{1+1}x^{1+1}+\frac{1}{-2+1}x^{-2+1}]^2_1\\\frac{1}{2}x^{2}+\frac{1}{-1}x^{-1}]^2_1\\\frac{1}{2}x^{2}-\frac{1}{x}]^2_1\\\frac{1}{2}2^{2}-\frac{1}{2}-(\frac{1}{2}1^{2}-\frac{1}{1})\\\frac{1}{2}4-\frac{1}{2}-(\frac{1}{2}-1)\\2-\frac{1}{2}-\frac{1}{2}+1\\2-1+1\\\boxed{2}$

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