What is the integral from 0 to H of (3x-4)^{23} ?

The integral from 0 to H of (3x-4)^{23} is

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integral from 0 to H of (3x-4)^{23}

\int_{0}^{H} (3 x - 4)^{23} d x

\frac{2.41105 E 19 H ^{12} - 2.76334 E 19 H ^{9} + 2.0725 E 19 H ^{8} - 1.30039 E 19 H ^{7} + \frac{1.12737 E 19 H ^{24} - 3.60758 E 20 H ^{23} + 5.53162 E 21 H ^{22} - 5.4087 E 22 H ^{21} + 3.78609 E 23 H ^{20} - 2.01925 E 24 H ^{19} + 8.52571 E 24 H ^{18} - 2.9231 E 25 H ^{17} + 8.28212 E 25 H ^{16} - 1.96317 E 26 H ^{15} + 2.69151 E 26 H ^{6} - 1.13327 E 26 H ^{5} + 3.77755 E 25 H ^{4} - 9.59378 E 24 H ^{3} + 1.74432 E 24 H ^{2} - 2.0224 E 23 H + 11 ( 3.5694 E 25 H ^{14} + 1.12811 E 26 H ^{10} ) - 6.66287 E 26 H ^{13} - 1.18451 E 27 H ^{11}}{39916800}}{72}

Solution steps

\int_{0}^{H} (3 x - 4)^{23} d x

Apply u-substitution

= \int_{- 4}^{3 H - 4} \frac{u ^{23}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{- 4}^{3 H - 4} u^{23} d u

Apply the Power Rule

= \frac{1}{3} [\frac{u ^{24}}{24}]_{- 4}^{3 H - 4}

Compute the boundaries:

\frac{( 3 H - 4 ) ^{24}}{24} - \frac{2 ^{45}}{3}

= \frac{1}{3} (\frac{( 3 H - 4 ) ^{24}}{24} - \frac{2 ^{45}}{3})

Simplify

= \frac{2.41105 E 19 H ^{12} - 2.76334 E 19 H ^{9} + 2.0725 E 19 H ^{8} - 1.30039 E 19 H ^{7} + \frac{1.12737 E 19 H ^{24} - 3.60758 E 20 H ^{23} + 5.53162 E 21 H ^{22} - 5.4087 E 22 H ^{21} + 3.78609 E 23 H ^{20} - 2.01925 E 24 H ^{19} + 8.52571 E 24 H ^{18} - 2.9231 E 25 H ^{17} + 8.28212 E 25 H ^{16} - 1.96317 E 26 H ^{15} + 2.69151 E 26 H ^{6} - 1.13327 E 26 H ^{5} + 3.77755 E 25 H ^{4} - 9.59378 E 24 H ^{3} + 1.74432 E 24 H ^{2} - 2.0224 E 23 H + 11 ( 3.5694 E 25 H ^{14} + 1.12811 E 26 H ^{10} ) - 6.66287 E 26 H ^{13} - 1.18451 E 27 H ^{11}}{39916800}}{72}

\int \frac{1}{sec ^{4} ( x )} d x

s l o p e (4.4) (2.4)

\int \frac{sin ( x )}{x} d x

\int cos (\frac{π}{2}) x d x

x \to 2 lim (5 - x^{3})

What is the integral from 0 to H of (3x-4)^{23} ?
The integral from 0 to H of (3x-4)^{23} is