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