Math, asked by prashant247, 2 months ago

Solve any one of them
..
.
.
.
.​

Attachments:

Answers

Answered by factorialiscool
2

Answer:

\int _0^1e^ti+e^{-2t}j+tkdt=\frac{-j+e^2\left(j+k\right)}{2e^2}+i\left(-1+e\right)

Step-by-step explanation:

=\int _0^1e^tidt+\int _0^1e^{-2t}jdt+\int _0^1tkdt

=\int\limits^1_0 {e^tidt} =i(e-1)

=\int\limits^1_0 {e^-^2^tjdt}=\frac{e^2-1}{2e^2} j

=\int\limits^1_0 {tkdt}=k\frac{1}{2}

=i(e-1)+\frac{e^2-1}{2e^2} j+k\frac{1}{2}

\frac{e^2(j+k)-j}{2e^2} +(e-1)i in a composite image:

i(e-1)+\frac{e^2-1}{2e^2} j+k\frac{1}{2}

=\frac{-j+e^2(j+k)}{2e^2} +i(-1+e)\\

Answered by lakshvagela1234
1

Answer:

marks as brilliant please I hope my answer will help you

Attachments:
Similar questions