What is the integral from 1/2 to 3/2 of qsqrt(q-1/2) ?

The integral from 1/2 to 3/2 of qsqrt(q-1/2) is 11/15

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integral from 1/2 to 3/2 of qsqrt(q-1/2)

\int_{\frac{1}{2}}^{\frac{3}{2}} q q - \frac{1}{2} d q

\frac{11}{15}

Decimal Notation

0.73333 \dots

Solution steps

\int_{\frac{1}{2}}^{\frac{3}{2}} q q - \frac{1}{2} d q

Apply Integration By Parts

: [\frac{1}{2 2} (q^{2} 2 q - 1 - 22 \cdot \int \frac{q ^{2}}{2 2 2 q - 1} d q)]_{\frac{1}{2}}^{\frac{3}{2}}

= [\frac{1}{2 2} (q^{2} 2 q - 1 - 22 \cdot \int \frac{q ^{2}}{2 2 2 q - 1} d q)]_{\frac{1}{2}}^{\frac{3}{2}}

\int \frac{q ^{2}}{2 2 2 q - 1} d q = \frac{1}{30 2} (3 q^{2} + 2 q + 2) 2 q - 1

= [\frac{1}{2 2} (q^{2} 2 q - 1 - 22 \frac{1}{30 2} (3 q^{2} + 2 q + 2) 2 q - 1)]_{\frac{1}{2}}^{\frac{3}{2}}

Simplify

[\frac{1}{2 2} (q^{2} 2 q - 1 - 22 \frac{1}{30 2} (3 q^{2} + 2 q + 2) 2 q - 1)]_{\frac{1}{2}}^{\frac{3}{2}} : [\frac{1}{15 2} (6 q^{2} - q - 1) 2 q - 1]_{\frac{1}{2}}^{\frac{3}{2}}

= [\frac{1}{15 2} (6 q^{2} - q - 1) 2 q - 1]_{\frac{1}{2}}^{\frac{3}{2}}

Compute the boundaries:

\frac{11}{15}

= \frac{11}{15}

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What is the integral from 1/2 to 3/2 of qsqrt(q-1/2) ?
The integral from 1/2 to 3/2 of qsqrt(q-1/2) is 11/15