Q. 59 - Calculus

Question

Give a geometric argument to prove Theorem 4.13(b): For any real numbers $0 < a < b,$ 

 ${\int }_a^{b}xdx=\frac{1}{2}\left(b^2-a^2\right)$

(Hint: Use a trapezoid.)

Step-by-Step Solution

Answer

The theorem 4.13(b)  is proved. 

${\int }_a^{b}xdx=\frac{1}{2}\left(b^2-a^2\right)$

Step 1. Given Information

We are given a theorem, 

${\int }_a^{b}xdx=\frac{1}{2}\left(b^2-a^2\right)$

tep 2. Proving the theorem

The proof is done by using a trapezoid.

The limit is the area covered by the trapezoid.

The area of a trapezoid with heights a, b and width $b-a$ is,

$$\begin{gathered} \frac{a+b}{2}(b-a)=\frac{1}{2}\left(ab+b^2-ab-a^2\right) \\ =\frac{1}{2}\left(b^2-a^2\right) \end{gathered}$$

Hence Proved.

${\int }_a^{b}xdx=\frac{1}{2}\left(b^2-a^2\right)$