To determine

To evaluate:

The integral ∫14x2-4x+2 dx by using the form of the definition of the integral given in theorem (4)

Answer

-3

Explanation

1) Concept:

Theorem (4):

If f is integrable on [a, b] then

∫abfxdx=limn→∞⁡∑i=1nfxi ∆x

where ∆x= b-an and xi=a+i ∆x

f(x) is called as an integrand,n is the sub interval,a is the lower limit, and b is the upper limit.

2) Formula:

i)∑i=1ni2= nn+1(2n+1)6

ii)∑i=1ni= n(n+1)2

iii)∑i=1ncai=c∑i=1naiwhere c is a constant

iv)∑i=1n(ai-bi)=∑i=1nai-∑i=1nbi

v)∑i=1nc =nc

3) Given:

∫14x2-4x+2 dx

4) Calculation:

Here, a=1, b=4 and fx=x2-4x+2

Substituting value of a and b in ∆x,

∆x= b-an

∆x= 4-1n

∆x= 3n

Now find xi,

xi=a+i ∆x

xi=1+i 3n

xi=1+3in

By using theorem (4),

∫14x2-4x+2 dx=limn→∞⁡∑i=1nf1+3in3n

= limn→∞⁡3n∑i=1nf1+3in

Thus by using formula (iii),

∫14x2-4x+2 dx= limn→∞⁡3n∑i=1nf1+3in

=limn→∞⁡3n∑i=1n1+3in2-41+3in+2

=limn→∞⁡3n∑i=1n1+6in+9i2n2-4-12in+2

= limn→∞⁡3n∑i=1n9i2n2-6in-1

By using formula (iv),

= limn→∞⁡3n9n2∑i=1ni2-6n∑i=1ni-∑i=1n1

By using formula (i), (ii) and (v),

= limn→∞⁡3n9n2n(n+1)(2n+1)6-6nn(n+1)2-n

=limn→∞⁡3n3n(n+1)(2n+1)2-3(n+1)-n

By applying the distributive property,

=limn→∞⁡3n3(n+1)(2n+1)2n-3(n+1)-n

=limn→∞⁡9(n+1)(2n+1)2n2-9(n+1)n-3

=limn→∞⁡92n+1n2n+1n-9n+1n-3

=limn→∞⁡921+1n2+1n-91+1n-3

Now by using limit laws,

∫14x2-4x+2 dx=921+02+0-91+0-3

=9212-91-3

=9-9-3

= -3

Therefore, ∫14x2-4x+2 dx=-3

Conclusion:∫14x2-4x+2 dx=-3