Tolong bantuin yaa no. 3-4 nyaa

Tolong bantuin yaa no. 3-4 nyaa

$frac{x}{sqrt{3x^{2}+1}}$ ——————————– (1)

Let u(x) = $3x^{2}+1}$

jadi

du/dx = $6x$

dx= $frac{du}{6x}$ ———————- (2)

lalu cari upper "boundary valuenya"

u(1) = 4 ———————— (3)

u(0) = 1 ———————— (4)

lalu ganti dx menjadi $frac{du}{6x}$

dan ganti boundary value nya

dari begini

$intlimits^1_0 {frac{x}{sqrt{3x^{2}+1}}} , dx$

$intlimits^{u(1) = 4}_{u(0)=1} {frac{x}{sqrt{u}}frac{1}{6x} } , du$

$intlimits^{u(1) = 4}_{u(0)=1} {frac{1}{6sqrt{u}}} , du$

$frac{1}{6}intlimits^{u(1) = 4}_{u(0)=1} {frac{1}{sqrt{u}}} , du$

integral nya $frac{1}{sqrt{u} }$ itu 2 $sqrt{u}$

apply boundary limit

2 $sqrt{4}$ – 2 $sqrt{1}$

= $frac{1}{6}$ *(2*2 – 2*1) = $frac{1}{6}$ * 2 = $frac{2}{6}$ = $frac{1}{3}$