Answer 1

Here, a = 1, b = 3 and f(x)=e^2−3x+x²+1.

∴nh=b−a=3−1=2

We know

`int_a^b f(x)dx=lim_(h->0)h{f(a)+f(a+h)+f(a+2h+....+f[a+(n-1)h])}`

Now,

f(a)=f(1)=e^2−3×1+1²+1

f(a+h)=f(1+h)=e^2−3(1+h)+(1+h)²+1

f(a+2h)=f(1+2h)=e^2−3(1+2h)+(1+2h)²+1

.

.

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f[a+(n−1)h]=f[1+(n−1)h]=e^{2−3[1+(n−1)h]}+[1+(n−1)h]²+1

Adding these equations, we get

f(a)+f(a+h)+f(a+2h)+...+f[a+(n−1)h]=[e^2−3×1+1²+1]+[e^2−3×(1+h)+(1+h)²+1]

+[e^{2−3×(1+2h)}+(1+2h)2+1]+...

+{e^{2−3[1+(n−1)h]}+[1+(n−1)h]²+1}

`therefore int_1^3(e^(2-3x)+x^2+1)dx`

`=lim_(h->0)h[e^2.(e^(-3)+e^(-3(1+h))`

`+e^(-3(1+2h))+....+e^(3(1+(n-1)h)))]+lim_(h->0)h[1^2`

`+(1+h)^2+(1+2h)^2+.....+[1+(n-1)^2h]]+lim_(h->0)h[1+1+1+...+1]`

`=lim_(h->0)h{e^2xx(e^-3(1-e^(3nh))/(1-e^(-3h)))}`

`+lim_(h->0)h{(1+1+1+...+1)+2h[1+2+3+.....+(n-1)]`

`+h^2(1^2+2^2+3^2+.....+(n-1)^2)}lim_(h->0)h(1+1+1+...+1)`

`=lim_(h->0)h{(e^(-1)(1-e^(-6)))/(1-e^(-3h))}+lim_(h->0)h[n+2hxx((n-1)n)/2+h^2xx((n-1)n(2n-1))/6]+lim_(h->0)hn`

`=lim_(h->0)h{(e^(-1)(1-e^(-6)))/(1-e^(-3h))}+lim_(h->0)h[2nh+(nh-h)nh+((nh-1)nh(2nh-1))/6]`

`=1/e(1-1/e^6)xx(lim_(h->0)e^(3h))/(3xxlim_(h->0)((e^(3h)-1)/(3h)))+lim_(h->0)[2xx2+((2-h)xx2+(2-h)xx2xx(2xx2-h))/6]`

`=1/e(1-1/e^6)xx1/(3xx1)+(4+4+8/3)`

`=1/(3e)(1-1/3^6)+32/3`

Answer 2

Let f(x) = `int (e^(2-3x) + x^2 +1) dx`  `("where" lim_(x->3))`

Here h = `2/n`

`= lim_(h->0) h [f(1) + f(1 + h) + f(1 + 2h) + .... + f(1 + (n-1)h)]`   

`= lim_(h->0) h [(e^-1 + 2) + (e^(-1-3h) +2 + 2h + h^2) + ... (e^(-1-6h) + 2 + 4h + 4h^2) + .... + .....(e^(-1-3(n-1)h) + 2 + 2(n-1)h + (n-1)^2 + h^2)]`

`= lim_(h->0) h [e^-1 (1 + e^-3h + e^-6h + ......+e^(-3(n-1)h) + 2n + 2h (1 + 2+...+(n-1)) + h^2(1^2 +2^2 + .... + (n-1)^2)]`

`=lim_(h->0) h [e^1  (e^(-3nh) - 1)/(e^(-3n) - 1) .  + 2nh + 2 (nh (nh - n))/2 + (nh (nh - h)(2nh - h))/6]`

`= e^-1  (e^-6 - 1)/-3 + 4 + 4 + 8/3`

`= -e^-1  (e^-6 - 1)/3 + 32/3`