To find the numerical value of \[\int\limits_{ - 2}^2 {\left( {p{x^2} + qx + s} \right)dx} \], is it necessary to know the values of the constants?
A. \[p\]
B. \[q\]
C. \[s\]
D. \[p\] and \[s\]

Hint: Here, a definite integral is given. First, separate the terms present in the given integral. Then, check whether the functions present in the integral are an even or an odd function. If the function is odd, then the value of that integral is 0. If the function is even, then solve the integral by applying the integration rule \[\int\limits_{ - a}^a {f\left( x \right)} dx = 2\int\limits_0^a {f\left( x \right)} dx\]. In the end, simplify the integral to get the required answer.



Formula Used:\[\int\limits_{ - a}^a {f\left( x \right) dx} = 0\], if the function \[f\left( x \right)\] is an odd function. Means, \[f\left( { - x} \right) = - f\left( x \right)\]
\[\int\limits_{ - a}^a {f\left( x \right) dx} = 2\int\limits_0^a {f\left( x \right) dx} \], if the function \[f\left( x \right)\] is an even function. Means, \[f\left( { - x} \right) = f\left( x \right)\]
\[\int\limits_a^b {{x^n}} dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]_a^b\]
\[\int\limits_a^b n dx = \left[ {nx} \right]_a^b\], where \[n\] is a number.



Complete step by step solution:The given integral is \[\int\limits_{ - 2}^2 {\left( {p{x^2} + qx + s} \right)dx} \].
Let consider,
\[I = \int\limits_{ - 2}^2 {\left( {p{x^2} + qx + s} \right)dx} \]
Simplify the integral by applying the sum rule of the integration \[\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]} dx = \int\limits_a^b {f\left( x \right)} dx + \int\limits_a^b {g\left( x \right)} dx\].
\[I = \int\limits_{ - 2}^2 {\left( {p{x^2} + s} \right)dx} + \int\limits_{ - 2}^2 {qx} dx\]
Let consider,
\[f\left( x \right) = p{x^2} + s\]
Now we have to check whether the above function is an odd function or an even function.
So, let’s calculate the value of \[f\left( { - x} \right)\].
\[f\left( { - x} \right) = p{\left( { - x} \right)^2} + s\]
\[ \Rightarrow f\left( { - x} \right) = p{x^2} + s\]
\[ \Rightarrow f\left( { - x} \right) = f\left( x \right)\]
Therefore, the function \[f\left( x \right) = p{x^2} + s\] is an even function.

Again consider, \[g\left( x \right) = qx\]
let’s calculate the value of \[g\left( { - x} \right)\].
\[g\left( { - x} \right) = q\left( { - x} \right)\]
\[ \Rightarrow g\left( { - x} \right) = - qx\]
\[ \Rightarrow g\left( { - x} \right) = - g\left( x \right)\]
Therefore, the function \[g\left( x \right) = qx\] is an odd function.

Now apply the property of the definite integral \[\int\limits_{ - a}^a {f\left( x \right) dx} = 0\], if the function \[f\left( x \right)\] is an odd function and \[\int\limits_{ - a}^a {f\left( x \right) dx} = 2\int\limits_0^a {f\left( x \right) dx} \], if the function \[f\left( x \right)\] is an even function.
We get,
\[I = 2\int\limits_0^2 {\left( {p{x^2} + s} \right)dx} + 0\]
\[ \Rightarrow I = 2\int\limits_0^2 {\left( {p{x^2} + s} \right)dx} \]
Solve the integral by applying the integration formula \[\int\limits_a^b {{x^n}} dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]_a^b\] and \[\int\limits_a^b n dx = \left[ {nx} \right]_a^b\].
 \[ \Rightarrow I = 2\left[ {p\dfrac{{{x^3}}}{3} + sx} \right]_0^2\]
Apply the upper and lower limits.
\[ \Rightarrow I = 2\left[ {\left[ {p\dfrac{{{2^3}}}{3} + s\left( 2 \right)} \right] - \left[ {p\dfrac{{{0^3}}}{3} + s\left( 0 \right)} \right]} \right]\]
\[ \Rightarrow I = 2\left[ {\dfrac{{8p}}{3} + 2s} \right]\]
\[ \Rightarrow I = \dfrac{{16p}}{3} + 4s\]
\[ \Rightarrow I = \dfrac{4}{3}\left( {4p + 3s} \right)\]
Therefore, \[\int\limits_{ - 2}^2 {\left( {p{x^2} + qx + s} \right)dx} = \dfrac{4}{3}\left( {4p + 3s} \right)\].
Thus, to calculate the numerical value of the integral it is necessary to know the values of \[p\] and \[s\].



Option ‘D’ is correct



Note: Students often do mistake to integrating \[\int\limits_a^b {{x^n}} dx\] . They apply the formula \[\int\limits_a^b {{x^n}} dx = \left[ {{x^{n + 1}}} \right]_a^b\] which is an incorrect formula. The correct formula is \[\int\limits_a^b {{x^n}} dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]_a^b\].