What is the integral from-1 to 2 of x^3sqrt(x^4+3) ?

The integral from-1 to 2 of x^3sqrt(x^4+3) is (19sqrt(19)-8)/6

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integral from-1 to 2 of x^3sqrt(x^4+3)

\int_{- 1}^{2} x^{3} x^{4} + 3 d x

\frac{19 19 - 8}{6}

Decimal

12.46984 \dots

Solution steps

\int_{- 1}^{2} x^{3} x^{4} + 3 d x

Apply u-substitution

= \int_{4}^{19} \frac{u}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{4}^{19} u d u

Apply the Power Rule

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{19}

Compute the boundaries:

\frac{38 19}{3} - \frac{16}{3}

= \frac{1}{4} (\frac{38 19}{3} - \frac{16}{3})

Simplify

= \frac{19 19 - 8}{6}

What is the integral from-1 to 2 of x^3sqrt(x^4+3) ?
The integral from-1 to 2 of x^3sqrt(x^4+3) is (19sqrt(19)-8)/6