What is the integral from 0 to 1 of (10sqrt(θ))/((1+θ^{3/2))^2} ?

The integral from 0 to 1 of (10sqrt(θ))/((1+θ^{3/2))^2} is 10/3

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integral from 0 to 1 of (10sqrt(θ))/((1+θ^{3/2))^2}

\int_{0}^{1} \frac{10 θ}{( 1 + θ ^{\frac{3}{2}} ) ^{2}} d θ

\frac{10}{3}

Decimal Notation

3.33333 \dots

Solution steps

\int_{0}^{1} \frac{10 θ}{( 1 + θ ^{\frac{3}{2}} ) ^{2}} d θ

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 10 \cdot \int_{0}^{1} \frac{θ}{( 1 + θ ^{\frac{3}{2}} ) ^{2}} d θ

Apply u-substitution

: \int_{1}^{2} \frac{2}{3 u ^{2}} d u

= 10 \cdot \int_{1}^{2} \frac{2}{3 u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 10 \cdot \frac{2}{3} \cdot \int_{1}^{2} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{1}^{2}

= 10 \cdot \frac{2}{3} [- \frac{1}{u}]_{1}^{2}

Simplify

10 \cdot \frac{2}{3} [- \frac{1}{u}]_{1}^{2} : \frac{20}{3} [- \frac{1}{u}]_{1}^{2}

= \frac{20}{3} [- \frac{1}{u}]_{1}^{2}

Compute the boundaries:

\frac{1}{2}

= \frac{20}{3} \cdot \frac{1}{2}

Simplify

= \frac{10}{3}

\frac{d}{d x} (\frac{x}{( x ^{2} - 1 ) ^{2}})

f (x) = \frac{e ^{x}}{x ^{6}}

\int 4 t^{3} t^{2} + 1 d t

y^{^{'}} = \frac{2 y}{x} - x^{2} y^{2}

x \to \infty lim (x^{s i n (x)})

What is the integral from 0 to 1 of (10sqrt(θ))/((1+θ^{3/2))^2} ?
The integral from 0 to 1 of (10sqrt(θ))/((1+θ^{3/2))^2} is 10/3