Question

Let \[a,b,c\] be non-zero real numbers such that \[\int\limits_0^3 {\left( {3a{x^2} + 2bx + c} \right)} dx = \int\limits_1^3 {\left( {3a{x^2} + 2bx + c} \right)} dx\], then which of the following equation is true?
A. \[a + b + c = 3\]
B. \[a + b + c = 1\]
C. \[a + b + c = 0\]
D. \[a + b + c = 2\]

Answer 1

Hint: Here, an equation of the definite integrals is given. First, simplify the left-hand side by applying the integration rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^c {f\left( x \right)} dx + \int\limits_c^b {f\left( x \right)} dx\]. Then, simplify the equation and cancel out the common terms from both sides. After that, solve the integrals by using the integration formulas \[\int\limits_a^b {ndx = \left[ {nx} \right]} _a^b = n\left( {b - a} \right)\] and \[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\]. In the end, apply the limits and solve the equation to get the required answer.



Formula Used:\[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^c {f\left( x \right)} dx + \int\limits_c^b {f\left( x \right)} dx\]
\[\int\limits_a^b {ndx = \left[ {nx} \right]} _a^b = n\left( {b - a} \right)\]
\[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\]



Complete step by step solution:The given equation of the definite integral is \[\int\limits_0^3 {\left( {3a{x^2} + 2bx + c} \right)} dx = \int\limits_1^3 {\left( {3a{x^2} + 2bx + c} \right)} dx\].
Let’s simplify the left-hand side by applying the integration rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^c {f\left( x \right)} dx + \int\limits_c^b {f\left( x \right)} dx\].
We get,
\[\int\limits_0^1 {\left( {3a{x^2} + 2bx + c} \right)} dx + \int\limits_1^3 {\left( {3a{x^2} + 2bx + c} \right)} dx = \int\limits_1^3 {\left( {3a{x^2} + 2bx + c} \right)} dx\]
Cancel out the common terms from both sides.
\[\int\limits_0^1 {\left( {3a{x^2} + 2bx + c} \right)} dx = 0\]
Now solve the integral by using the integration formulas \[\int\limits_a^b {ndx = \left[ {nx} \right]} _a^b = n\left( {b - a} \right)\] and \[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\].
We get,
\[\left[ {\dfrac{{3a{x^3}}}{3} + \dfrac{{2b{x^2}}}{2} + cx} \right]_0^1 = 0\]
\[ \Rightarrow \left[ {a{x^3} + b{x^2} + cx} \right]_0^1 = 0\]
Apply the upper and lower limits.
\[ \Rightarrow a\left( {{1^3} - {0^3}} \right) + b\left( {{1^2} - {0^2}} \right) + c\left( {1 - 0} \right) = 0\]
\[ \Rightarrow a\left( {1 - 0} \right) + b\left( {1 - 0} \right) + c\left( 1 \right) = 0\]
\[ \Rightarrow a + b + c = 0\]



Option ‘C’ is correct



Note: Students get confused and solve the integral \[\int {{x^n}dx = {x^{n + 1}}} \], which is an incorrect formula. They forget to divide it by \[n + 1\]. The correct formula is \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} \].