int_0^{16} frac{sqrt{x}}{1+sqrt{x}} d x= | vedclass

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Question

(C) Let $I = \int_0^{16} \frac{\sqrt{x}}{1+\sqrt{x}} dx$. By performing algebraic manipulation, we can write the integrand as: $\frac{\sqrt{x}}{1+\sqr

Vedclass · Accepted Answer

$8+2 \log 5$

Vedclass · Answer

(C) Let $I = \int_0^{16} \frac{\sqrt{x}}{1+\sqrt{x}} dx$. By performing algebraic manipulation, we can write the integrand as: $\frac{\sqrt{x}}{1+\sqrt{x}} = \frac{\sqrt{x}+1-1}{1+\sqrt{x}} = 1 - \frac{1}{1+\sqrt{x}}$. Thus, $I = \int_0^{16} 1 dx - \int_0^{16} \frac{1}{1+\sqrt{x}} dx = 16 - \int_0^{16} \frac{1}{1+\sqrt{x}} dx$. To evaluate $\int_0^{16} \frac{1}{1+\sqrt{x}} dx$, substitute $t = 1+\sqrt{x}$, so $\sqrt{x} = t-1$ and $x = (t-1)^2$, which gives $dx = 2(t-1) dt$. When $x=0$, $t=1$. When $x=16$, $t=1+\sqrt{16}=5$. So, $\int_0^{16} \frac{1}{1+\sqrt{x}} dx = \int_1^5 \frac{2(t-1)}{t} dt = 2 \int_1^5 (1 - \frac{1}{t}) dt = 2 [t - \ln|t|]_1^5$. $= 2 [(5 - \ln 5) - (1 - \ln 1)] = 2 [4 - \ln 5] = 8 - 2 \ln 5$. Therefore, $I = 16 - (8 - 2 \ln 5) = 8 + 2 \ln 5$.

$\int_0^{16} \frac{\sqrt{x}}{1+\sqrt{x}} d x=$

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