- #1
tmt1
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- 0
I have:
$$\int_{1}^{3} \frac{1}{\sqrt{3 - x}} \,dx$$
I can do $u = \sqrt{3 -x}$, and $du = \frac{1}{2 \sqrt{3 - x}} dx $ and $dx = 2 \sqrt{3 - x} du $. Plug into original equation:
$$\int_{1}^{3} \frac{2 u }{u} \,du$$ and $2 \int_{1}^{3} \,du = 2u = 2 \sqrt{3 - x} + C$
So $(2\sqrt{3 - 3}) - (2\sqrt{3 - 1})$ Or $- 2\sqrt{2}$. Is this correct?
$$\int_{1}^{3} \frac{1}{\sqrt{3 - x}} \,dx$$
I can do $u = \sqrt{3 -x}$, and $du = \frac{1}{2 \sqrt{3 - x}} dx $ and $dx = 2 \sqrt{3 - x} du $. Plug into original equation:
$$\int_{1}^{3} \frac{2 u }{u} \,du$$ and $2 \int_{1}^{3} \,du = 2u = 2 \sqrt{3 - x} + C$
So $(2\sqrt{3 - 3}) - (2\sqrt{3 - 1})$ Or $- 2\sqrt{2}$. Is this correct?