The correct option is A True
We have to state whether ∫20exdx can be represented as 2limn→∞1n⎡⎢
⎢⎣e0+e2n+e4n+...+e2(n−1)n⎤⎥
⎥⎦ is true or false.
We know that ∫baf(x)dx=limn→∞[b−an∑nk=1f(a+kb−an)]
Here we have f(x)=ex,a=0,b=2
Therefore ∫20exdx=limn→∞[2−0n∑nk=1f(0+k2−0n)]
=limn→∞[2n∑nk=1f(2kn)]
=2limn→∞[1n∑n−1k=0f(2kn)]
=2limn→∞1n[f(0)+f(2n)+f(4n)+...+f(2(n−1)n)]
Since f(x)=ex we get
∫20exdx=2limn→∞1n⎡⎢
⎢⎣e0+e2n+e4n+...+e2(n−1)n⎤⎥
⎥⎦
Hence the answer is TRUE.