Question

\(\int\limits_{ - 5}^5 {(a{x^2} + bx + c)dx = 2\int\limits_0^5 {(a{x^2} + c)dx} } \)

Short answer

The answer is TRUE.

Step-by-step solution

Solving the LHS

\(\begin{array}{l}\int\limits_{ - 5}^5 {(a{x^2} + bx + c)dx = \left( {a\frac{{{x^3}}}{3} + b\frac{{{x^2}}}{2} + cx} \right)_{ - 5}^5} \\ = \left( {a\frac{{{5^3}}}{3} + b\frac{{{5^2}}}{2} + c5} \right) - \left( {a\frac{{{{( - 5)}^3}}}{3} + b\frac{{{{( - 5)}^2}}}{2} + c( - 5)} \right)\end{array}\)

\(\begin{array}{l} = 2\left( {a\frac{{{5^3}}}{3} + c(5)} \right)\\ = \frac{{250}}{3}a + 10c\end{array}\)

Step 2: Solving the RHS

\(\begin{array}{l}2\int\limits_0^5 {(a{x^2} + c)dx = 2\left( {a\frac{{{x^3}}}{3} + cx} \right)_0^5} \\ = 2\left( {\left( {a\frac{{{5^3}}}{3} + c5} \right) - \left( {a\frac{{{0^3}}}{3} + c.0} \right)} \right)\end{array}\)

\(\begin{array}{l} = 2\left( {a\frac{{{5^3}}}{3} + c5} \right)\\ = \frac{{250}}{3}a + 10c\end{array}\)

Step 3: Final Answer from above two steps

Hence it can be shown that

\(\int\limits_{ - 5}^5 {(a{x^2} + bx + c)dx = 2\int\limits_0^5 {(a{x^2} + c)dx} } \)

Hence it is true.