# Solved Problems Using Definition of Definite Integrals

Here we solve few problems of definite integrals.

Problem 1:  By the definition of definite integrals, find int_a^b dx.

Solution: From the definition of definite integrals, we have

int_a^b dx=lim_{h to 0} h sum_{1}^{n} 1, quad where  nh=b-a

=lim_{h to 0} h (1+1+1+ cdots text{till} n text{-terms})

=lim_{h to 0} h cdot n

=lim_{h to 0} b-a = b-a

Problem 2:  By the definition of definite integrals, find int_0^1 x dx.

Solution: From the definition of definite integrals, we have

int_0^1 x dx=lim_{h to 0} h sum_{r=1}^{n} rh, quad where  nh=1-0=1

=lim_{h to 0} h (h+2h+3h+ cdots nh)

=lim_{h to 0} h^2 (1+2+3+ cdots n)

=lim_{h to 0} h^2 frac{n(n+1)}{2}

[Formula used: 1+2+cdots+n=frac{n(n+1)}{2}]

=lim_{h to 0}frac{nh(nh+h)}{2}

=lim_{h to 0}frac{1(1+h)}{2}

=frac{1(1+0)}{2}=frac{1}{2}

Problem 3:  By the definition of definite integrals, find int_1^2 x dx.

Solution: From the definition of definite integrals, we have

int_1^2 x dx=lim_{h to 0} h sum_{r=1}^{n} (1+rh), quad where  nh=2-1=1

=lim_{h to 0} h [(1+h)+(1+2h)+cdots+(1+nh)]

=lim_{h to 0} h [(1+1+ cdots text{till} n text{-terms})+ h(1+2+cdots n)]

=lim_{h to 0}   h[n+h frac{n(n+1)}{2}]

[Formula used: 1+2+cdots+n=frac{n(n+1)}{2}]

=lim_{h to 0}[nh+frac{nh(nh+h)}{2}]

=lim_{h to 0}[1+frac{1(1+h)}{2}] quadas  nh=1

=1+frac{1(1+0)}{2}=frac{3}{2}

Problem 4:  By the definition of definite integrals, find int_0^1 3x^2 dx.

Solution: From the definition of definite integrals, we have

int_0^1 3x^2 dx=lim_{h to 0} h sum_{r=1}^{n} 3(rh)^2, quad where  nh=1-0=1

=lim_{h to 0} h cdot 3h^2 sum_{r=1}^{n} r^2

=lim_{h to 0} 3h^3 (1^2+2^2+3^2+ cdots n^2)

=lim_{h to 0} 3h^3 cdot frac{n(n+1)(2n+1)}{6}

[Formula used: 1^2+2^2+cdots+n^2=frac{n(n+1)(2n+1)}{6}]

=3lim_{h to 0}frac{nh(nh+h)(2nh+h)}{6}

=3lim_{h to 0}frac{1(1+h)(2+h)}{6} quad as  nh=1

=3frac{1(1+0)(2+0)}{6}=1