What is the integral from 15 to 30 of x^2sin(5)x^3 ?

The integral from 15 to 30 of x^2sin(5)x^3 is sin(5)(((30)^6)/6-((15)^6)/6)

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integral from 15 to 30 of x^2sin(5)x^3

\int_{1 5^{\circ}}^{3 0^{\circ}} x^{2} sin (5) x^{3} d x

sin (5) (\frac{( 3 0 ^{\circ} ) ^{6}}{6} - \frac{( 1 5 ^{\circ} ) ^{6}}{6})

Decimal Notation

- 0.00324 \dots

Solution steps

\int_{1 5^{\circ}}^{3 0^{\circ}} x^{2} sin (5) x^{3} d x

Take the constant out:

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

= sin (5) \cdot \int_{1 5^{\circ}}^{3 0^{\circ}} x^{2} x^{3} d x

Simplify

x^{2} x^{3} : x^{5}

= sin (5) \cdot \int_{1 5^{\circ}}^{3 0^{\circ}} x^{5} d x

Apply the Power Rule

: [\frac{x ^{6}}{6}]_{1 5^{\circ}}^{3 0^{\circ}}

= sin (5) [\frac{x ^{6}}{6}]_{1 5^{\circ}}^{3 0^{\circ}}

Compute the boundaries:

\frac{( 3 0 ^{\circ} ) ^{6}}{6} - \frac{( 1 5 ^{\circ} ) ^{6}}{6}

= sin (5) (\frac{( 3 0 ^{\circ} ) ^{6}}{6} - \frac{( 1 5 ^{\circ} ) ^{6}}{6})

\int_{- 1}^{1} \frac{x}{2} d x

\int_{- 1}^{- 2} f (t)

\int_{0}^{1} [x^{\frac{1}{n}} - x] d x

\int_{0}^{2} x^{2} 4 x^{2} + 1 d x

\int_{0}^{x} e^{5 t^{2}} d t

What is the integral from 15 to 30 of x^2sin(5)x^3 ?
The integral from 15 to 30 of x^2sin(5)x^3 is sin(5)(((30)^6)/6-((15)^6)/6)