Use the Comparison Theorem to determine whether the integral is conver

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Use the Comparison Theorem to determine whether the integral is convergent or divergent. ∫0π2 sin2⁡(x)xdx divergent, because 2 sin2⁡(x)x > 2 x and ∫0π2 xdx is divergent. convergent, because 2 sin2⁡(x)x < 2 x and ∫0π2 xdx is convergent. divergent, because 2 sin2⁡(x)x < 2 x and ∫0π2 xdx is divergent. convergent, because 2 sin2⁡(x)x > 2 x and ∫0π2 xdx is convergent.

Use the Comparison Theorem to determine whether the integral is convergent or divergent. ∫0π2 sin2⁡(x)xdx divergent, because 2 sin2⁡(x)x > 2 x and ∫0π2 xdx is divergent. convergent, because 2 sin2⁡(x)x < 2 x and ∫0π2 xdx is convergent. divergent, because 2 sin2⁡(x)x < 2 x and ∫0π2 xdx is divergent. convergent, because 2 sin2⁡(x)x > 2 x and ∫0π2 xdx is convergent.

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Use the Comparison Theorem to determine whether the integral is convergent or divergent.

\int_{0}^{π} \frac{2 \sin^{2} (x)}{\sqrt{x}} d x

divergent, because

\frac{2 \sin^{2} (x)}{\sqrt{x}} > \frac{2}{\sqrt{x}}

and

\int_{0}^{π} \frac{2}{\sqrt{x}} d x

is divergent. convergent, because

\frac{2 \sin^{2} (x)}{\sqrt{x}} < \frac{2}{\sqrt{x}}

and

\int_{0}^{π} \frac{2}{\sqrt{x}} d x

is convergent. divergent, because

\frac{2 \sin^{2} (x)}{\sqrt{x}} < \frac{2}{\sqrt{x}}

and

\int_{0}^{π} \frac{2}{\sqrt{x}} d x

is divergent. convergent, because

\frac{2 \sin^{2} (x)}{\sqrt{x}} > \frac{2}{\sqrt{x}}

and

\int_{0}^{π} \frac{2}{\sqrt{x}} d x

is convergent.

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