Mathematics

$$\int \left( x ^ { 6 } + 7 x ^ { 5 } + 6 x ^ { 4 } + 5 x ^ { 3 } + 4 x ^ { 2 } + 3 x + 1 \right) e ^ { x } d x$$ equals


ANSWER

$$\sum _ { i = 1 } ^ { 6 } x ^ { i } e ^ { x } + c$$


SOLUTION
$$\displaystyle\int (x^6+7x^5+6x^4+5x^3+4x^2+3x+1)e^xdx$$
$$=\displaystyle\int (x^6+6x^5)e^xdx+\displaystyle\int (x^5+5x^4)e^xdx+\displaystyle\int (x^4+4x^3)e^xdx+\displaystyle\int (x^3+3x^2)e^xdx=\displaystyle\int (x^2+2x)e^xdx+\displaystyle\int (x+1)e^xdx$$.
$$\displaystyle\int [f(x)+f'(x)]e^xdx=f(x)e^x$$
Using this fact, we get
$$\displaystyle\int (x^6+6x^5)e^xdx=\displaystyle\int (x^6+(x^6)')e^xdx$$
$$=x^6\cdot e^xdx$$
Hence G.E$$=x^6e^x+x^5e^x+...…+xe^x+c$$
$$=\displaystyle\sum^6_{i=1}x^ie^x+c$$.
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Single Correct Medium Published on 17th 09, 2020
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