Question

Integrate the following:
int(2t -4)^4 dt = please give detailed answer​

Answer 1

Step-by-step explanation:

We have,

$\int(2t - 4)^{4} dt$

$= \int \{ (2t)^{4} - (2t)^{3}(4) + (2t)^{2} {(4)}^{2} - (2t) {(4)}^{3} + {(4)}^{4} \} dt$

$= \int \{16 (t)^{4} - 32(t)^{3} + 64(t)^{2} - 128(t) + 256 \} dt$

$= \int 16 (t)^{4} dt- \int32(t)^{3}dt + \int64(t)^{2}dt - \int 128(t) dt + \int256 dt$

$=16 \int (t)^{4} dt- 32 \int(t)^{3}dt + 64\int(t)^{2}dt - 128\int (t) dt + 256 \int dt$

$=16 \frac{t^{5}}{5} - 32 \frac{t^{4}}{4} + 64\frac{t^{3}}{3} - 128\frac{( {t}^{2} }{2} + 256t + C$

$= \frac{16}{5} {t}^{5} - \frac{32}{4} t^{4}+ \frac{64}{3} t^{3}- \frac{ 128 }{2}{t}^{2} + 256t + C$

$= \frac{16}{5} {t}^{5} - 8 t^{4}+ \frac{64}{3} t^{3}- 64{t}^{2} + 256t + C$