Hasil dari.... Question from @Hutami127

September 2019 1 40 Report

Hasil dari $\int\limits^2_0 {x^{2}\sqrt{4-x^{2}}} \, dx$

Verified answer

$\displaystyle \int x^2\sqrt{4-x^2}\,dx=\int(4+x^2-4)\sqrt{4-x^2}\,dx\\\int x^2\sqrt{4-x^2}\,dx=\int(4)\sqrt{4-x^2}+(x^2-4)\sqrt{4-x^2}\,dx\\\int x^2\sqrt{4-x^2}\,dx=\int4\sqrt{4-x^2}\,dx-\int(4-x^2)\sqrt{4-x^2}\,dx\\\int x^2\sqrt{4-x^2}\,dx=\int4(4-x^2)^\frac12\,dx-\int(4-x^2)^\frac32\,dx$

$\displaystyle \text{misal:}\\x=2\sin u\\u=\arcsin\frac12x\\dx=2\cos u\,du\\\\\\\int4(4-x^2)^\frac12\,dx=\int4(4-4\sin^2u)^\frac12\cdot2\cos u\,du\\\int4(4-x^2)^\frac12\,dx=\int4(4\cos^2u)^\frac12\cdot2\cos u\,du\\\int4(4-x^2)^\frac12\,dx=\int16\cos^2u\,du\\\int4(4-x^2)^\frac12\,dx=\int16\cdot\frac12(\cos2u+1)\,du\\\int4(4-x^2)^\frac12\,dx=8\cdot\frac12\sin2u+8u\\\int4(4-x^2)^\frac12\,dx=4\sin2u+8u$

$\displaystyle \int(4-x^2)^\frac32\,dx=\int(4-4\sin^2u)^\frac32\cdot2\cos u\,du\\\int(4-x^2)^\frac32\,dx=\int(4\cos^2u)^\frac32\cdot2\cos u\,du\\\int(4-x^2)^\frac32\,dx=\int16\cos^4u\,du\\\int(4-x^2)^\frac32\,dx=\int16(\cos^2u)^2\,du\\\int(4-x^2)^\frac32\,dx=\int16(\frac12(\cos2u+1))^2\,du\\\int(4-x^2)^\frac32\,dx=\int4(\cos^22u+2\cos2u+1)\,du\\\int(4-x^2)^\frac32\,dx=\int4(\frac12(\cos4u+1)+2\cos2u+1)\,du\\\int(4-x^2)^\frac32\,dx=\int2\cos4u+8\cos2u+6\,du\\\int(4-x^2)^\frac32\,dx=\frac12\sin4u+4\sin2u+6u$

$\displaystyle \int x^2\sqrt{4-x^2}\,dx=\int4(4-x^2)^\frac12\,dx-\int(4-x^2)^\frac32\,dx\\\int x^2\sqrt{4-x^2}\,dx=4\sin2u+8u-\frac12\sin4u-4\sin2u-6u\\\int x^2\sqrt{4-x^2}\,dx=2u-\sin2u\cos2u\\\int x^2\sqrt{4-x^2}\,dx=2u-2\sin u\cos u(2\cos^2u-1)\\\int x^2\sqrt{4-x^2}\,dx=2u-4\sin u\cos^3u+2\sin u\cos u\\\int x^2\sqrt{4-x^2}\,dx=2\arcsin\frac12x-2x\sqrt{\left(1-\frac{x^2}4\right)^3}+x\sqrt{1-\frac{x^2}{4}}$

$\displaystyle \int\limits^2_0x^2\sqrt{4-x^2}\,dx=\left2\arcsin\frac12x-2x\sqrt{\left(1-\frac{x^2}4\right)^3}+x\sqrt{1-\frac{x^2}{4}}\right|^2_0\\\int\limits^2_0x^2\sqrt{4-x^2}\,dx=2\arcsin\frac12(2)-2(2)\sqrt{\left(1-\frac{2^2}4\right)^3}+2\sqrt{1-\frac{2^2}{4}}-2\arcsin\frac12(0)-2(0)\sqrt{\left(1-\frac{0^2}4\right)^3}+0\sqrt{1-\frac{0^2}{4}}$
$\displaystyle \int\limits^2_0x^2\sqrt{4-x^2}\,dx=\pi-0+0-0\\\boxed{\boxed{\int\limits^2_0x^2\sqrt{4-x^2}\,dx=\pi}}$

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