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Question: How do you evaluate the integral of \(\int{(3-2x)dx}\) from -1 to 3?...

How do you evaluate the integral of ∫(3−2x)dx\int{(3-2x)dx} from -1 to 3?

Explanation

Solution

To evaluate the integral of a polynomial, we need to consider the inverse of the power rule used in differentiation, i.e., ∫xndx=xn+1n+1+c\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c . If the integral value of the function is within the given bounds, we subtract the value of the function at the lower bound from the function at the higher bound.

Complete Step by Step Solution:
Considering the lower and upper limits the question can be written as
=∫−13(3−2x)dx=\int\limits_{-1}^{3}{(3-2x)dx}
Expanding the given integral, we get
=∫−133dx−∫−132xdx=\int\limits_{-1}^{3}{3dx-\int\limits_{-1}^{3}{2xdx}}
Integrating the above integral,
=3x−2(x22)∣−13=\left. 3x-2\left( \dfrac{{{x}^{2}}}{2} \right) \right|_{-1}^{3}
Simplifying,
=3x−(x2)∣−13=\left. 3x-\left( {{x}^{2}} \right) \right|_{-1}^{3}
Applying the bounds and substituting x values,
=[3(3)−(3)2]−[3(−1)−(−1)2]=[9−9]−[−3−1]=\left[ 3(3)-{{(3)}^{2}} \right]-\left[ 3(-1)-{{(-1)}^{2}} \right]=\left[ 9-9 \right]-\left[ -3-1 \right]
Simplifying the above expression, we get
=[0]−[−4]=4=[0]-[-4]=4

Therefore, ∫−13(3−2x)dx=4\int\limits_{-1}^{3}{(3-2x)dx}=4

Note:
Alternate Method:
Taking the lower and upper limits into consideration the question can be written as
=∫−13(3−2x)dx=\int\limits_{-1}^{3}{(3-2x)dx}
Using the formula, ∫abf(x)dx=lim⁡n→∞ ∑i=1nf(xi)Δx\int\limits_{a}^{b}{f(x)dx}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{f({{x}_{i}})\Delta x}
In the above formula, for each positive integer n, we have Δx=b−an\Delta x=\dfrac{b-a}{n} and for i=1,2,3,.......,n,i=1,2,3,.......,n, we get xi=a+iΔx{{x}_{i}}=a+i\Delta x .
For every n, Δx=b−an=3−(−1)n=4n\Delta x=\dfrac{b-a}{n}=\dfrac{3-(-1)}{n}=\dfrac{4}{n}
For xi=a+iΔx{{x}_{i}}=a+i\Delta x we get the expression as xi=−1+i4n=−1+4in{{x}_{i}}=-1+i\dfrac{4}{n}=-1+\dfrac{4i}{n}
Using this, we can write the expression of f(xi)f\left( {{x}_{i}} \right) as f(xi)=3−2xi=3−2(−1+4in)=5−8inf\left( {{x}_{i}} \right)=3-2{{x}_{i}}=3-2\left( -1+\dfrac{4i}{n} \right)=5-\dfrac{8i}{n}
Using summation formulas for simplifying the above expression,
⇒∑i=1nf(xi)Δx=∑i=1n(5−8in)4n\Rightarrow \sum\limits_{i=1}^{n}{f({{x}_{i}})\Delta x}=\sum\limits_{i=1}^{n}{\left( 5-\dfrac{8i}{n} \right)\dfrac{4}{n}}
Multiplying each term in the bracket with 4n\dfrac{4}{n}
=∑i=1n(20n−32in2)=\sum\limits_{i=1}^{n}{\left( \dfrac{20}{n}-\dfrac{32i}{{{n}^{2}}} \right)}
=20n∑i=1n(1)+32n2∑i=1n(i)=\dfrac{20}{n}\sum\limits_{i=1}^{n}{\left( 1 \right)+\dfrac{32}{{{n}^{2}}}}\sum\limits_{i=1}^{n}{\left( i \right)}
Simplifying the summation in each of the terms as ∑i=1n(1)=n\sum\limits_{i=1}^{n}{\left( 1 \right)=n} and ∑i=1n(i)=n(n+1)2\sum\limits_{i=1}^{n}{\left( i \right)}=\dfrac{n(n+1)}{2} we can simplify as =20n[n]−32n2[n(n+1)2]=\dfrac{20}{n}[n]-\dfrac{32}{{{n}^{2}}}\left[ \dfrac{n(n+1)}{2} \right]
So, ∑i=1nf(xi)Δx=20−322[n(n+1)n2]\sum\limits_{i=1}^{n}{f({{x}_{i}})\Delta x}=20-\dfrac{32}{2}\left[ \dfrac{n(n+1)}{{{n}^{2}}} \right]
To solve the expression, we need to evaluate the limit as n→∞n\to \infty . This implies that we need to evaluate lim⁡n→∞ [n(n+1)n2]\underset{n\to \infty }{\mathop{\lim }}\,\left[ \dfrac{n(n+1)}{{{n}^{2}}} \right]. The numerator of this expression will expand to a polynomial which has a leading term n2{{n}^{2}}, therefore, the limit as n→∞n\to \infty is 1.
⇒n(n+1)n2=(nn)(n+1n)=(1)(nn+1n)=1+1n\Rightarrow \dfrac{n(n+1)}{{{n}^{2}}}=\left( \dfrac{n}{n} \right)\left( \dfrac{n+1}{n} \right)=(1)\left( \dfrac{n}{n}+\dfrac{1}{n} \right)=1+\dfrac{1}{n}
Simplifying and completing the integration, we get
lim⁡n→∞ ∑i=1nf(xi)Δx=20−322lim⁡n→∞ [n(n+1)n2]\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{f({{x}_{i}})\Delta x}=20-\dfrac{32}{2}\underset{n\to \infty }{\mathop{\lim }}\,\left[ \dfrac{n(n+1)}{{{n}^{2}}} \right]
Applying the limit and simplifying,
⇒20−16=4\Rightarrow 20-16=4
Therefore, ∫−13(3−2x)dx=4\int\limits_{-1}^{3}{(3-2x)dx}=4