Evaluate ∫1^3(e^(2-3x)+x^2+1)dx as a limit of sum. - Mathematics

Sum

Evaluate int_1^3(e^(2-3x)+x^2+1)dx  as a limit of sum.

Solution 1

Here, a = 1, b = 3 and f(x)=e23x+x2+1.

nh=ba=31=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)=e23×1+12+1

f(a+h)=f(1+h)=e23(1+h)+(1+h)2+1

f(a+2h)=f(1+2h)=e23(1+2h)+(1+2h)2+1

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f[a+(n1)h]=f[1+(n1)h]=e23[1+(n1)h]+[1+(n1)h]2+1

f(a)+f(a+h)+f(a+2h)+...+f[a+(n1)h]=[e23×1+12+1]+[e23×(1+h)+(1+h)2+1]

+[e23×(1+2h)+(1+2h)2+1]+...

+{e23[1+(n1)h]+[1+(n1)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

Solution 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

Concept: Definite Integral as the Limit of a Sum
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