Q. 80 - Calculus

Question

In this exercise you will use two different methods to prove that, for any real numbers a, b, c, and k, ${\int }_{-k}^k\left(a{x}^2+c\right)dx = 2{\int }_0^k\left(a{x}^2+c\right)dx.$

(a) Prove this equality by using a geometric argument that involves signed area.

(b) Now prove the equality a different way, by using an algebraic argument and the Fundamental Theorem of Calculus.

Step-by-Step Solution

Answer

Hence, proved.

Step 1. Given Information.

a, b, c, and k are real numbers and constant.

Step 2. Proof of part (a).

$$\begin{gathered} Let f\left(x\right) = a{x}^2+c. \\ Now we know that f\left(x\right) is an even function i.e., f(-x) = f\left(x\right). \\ So, \\ {\int }_{-k}^{k}f\left(x\right)dx = {\int }_{-k}^{0}f\left(x\right)dx +{\int }_0^{k}f\left(x\right)dx, \\ Let -x = t for the first term only, \\ dt = -dx, we get, \\ ={\int }_k^0-f(-t)dt + {\int }_0^{k}f\left(x\right)dx, \\ = {\int }_0^{k}f(-t)dt + {\int }_0^{k}f\left(x\right)dx, \left[Since, f\left(x\right) is an even function.\right] \\ = {\int }_0^{k}f\left(t\right)dt + {\int }_0^{k}f\left(x\right)dx, \\ = {\int }_0^{k}f\left(x\right)dx + {\int }_0^{k}f\left(x\right)dx = 2{\int }_0^{k}f\left(x\right)dx. \\ Hence, proved. \end{gathered}$$

Step 3. Proof of part (b).

$$\begin{gathered} Let f\left(x\right) = a{x}^2+c. \\ Now we know that f\left(x\right) is an even function i.e., f(-x) = f\left(x\right). \\ So, \\ {\int }_{-k}^{k}f\left(x\right)dx = {\int }_{-k}^{0}f\left(x\right)dx +{\int }_0^{k}f\left(x\right)dx, \\ Let -x = t for the first term only, \\ dt = -dx, we get, \\ ={\int }_k^0-f(-t)dt + {\int }_0^{k}f\left(x\right)dx, \\ = {\int }_0^{k}f(-t)dt + {\int }_0^{k}f\left(x\right)dx, \left[Since, f\left(x\right) is an even function.\right] \\ = {\int }_0^{k}f\left(t\right)dt + {\int }_0^{k}f\left(x\right)dx, \\ = {\int }_0^{k}f\left(x\right)dx + {\int }_0^{k}f\left(x\right)dx = 2{\int }_0^{k}f\left(x\right)dx. \\ Hence, proved. \end{gathered}$$