Question

Should

$\int_{1}^{2}x^2-5x+6 = \int_{3}^{4}x^2-5x+6 ?$

[Because it is an even and symmetrical function?]

• A. Yes
• B. No

Answer 1

Answer: (A)

Yes

Answer 2

The given question asks whether the definite integral of the function $f(x) = x^2 - 5x + 6$ from 1 to 2 is equal to its definite integral from 3 to 4, based on the premise that it is an even and symmetrical function.

1. Analyze the function for symmetry: The function $f(x) = x^2 - 5x + 6$ is a quadratic function, which represents a parabola. The axis of symmetry for a parabola $ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$. For $f(x) = x^2 - 5x + 6$, we have $a=1$ and $b=-5$. So, the axis of symmetry is $x = -\frac{(-5)}{2 \times 1} = \frac{5}{2} = 2.5$. A function $f(x)$ is symmetric about the line $x=a$ if $f(a-h) = f(a+h)$ for any real number $h$. Let's check this for $a=2.5$: $f(2.5-h) = (2.5-h)^2 - 5(2.5-h) + 6 = (6.25 - 5h + h^2) - (12.5 - 5h) + 6 = 6.25 - 5h + h^2 - 12.5 + 5h + 6 = h^2 - 0.25$. $f(2.5+h) = (2.5+h)^2 - 5(2.5+h) + 6 = (6.25 + 5h + h^2) - (12.5 + 5h) + 6 = 6.25 + 5h + h^2 - 12.5 - 5h + 6 = h^2 - 0.25$. Since $f(2.5-h) = f(2.5+h)$, the function $f(x)$ is indeed symmetric about the line $x=2.5$. Note: The term "even function" specifically refers to symmetry about the y-axis ($x=0$). While $f(x)$ is symmetric, it is not an "even function" in the standard definition.

2. Analyze the integration intervals for symmetry: The first interval is $[1, 2]$. The second interval is $[3, 4]$. Let's express these intervals relative to the axis of symmetry $x=2.5$. For $[1, 2]$: $1 = 2.5 - 1.5$ $2 = 2.5 - 0.5$ So, the interval is $[2.5 - 1.5, 2.5 - 0.5]$.

For $[3, 4]$: $3 = 2.5 + 0.5$ $4 = 2.5 + 1.5$ So, the interval is $[2.5 + 0.5, 2.5 + 1.5]$.

We can see that if we have an axis of symmetry at $x=a$, and two intervals are $[a-k_2, a-k_1]$ and $[a+k_1, a+k_2]$, then these intervals are symmetrically placed with respect to $x=a$.

3. Property of definite integrals for symmetric functions and intervals: If a function $f(x)$ is symmetric about $x=a$ (i.e., $f(a-h) = f(a+h)$), then for any real numbers $k_1$ and $k_2$: $\int_{a-k_2}^{a-k_1} f(x) dx = \int_{a+k_1}^{a+k_2} f(x) dx$ This property holds because the graph of the function over the interval $[a-k_2, a-k_1]$ is a mirror image of the graph over $[a+k_1, a+k_2]$ with respect to the line $x=a$. Consequently, the signed areas under the curve for these symmetric intervals will be equal.

4. Conclusion: Since the function $f(x) = x^2 - 5x + 6$ is symmetric about $x=2.5$, and the intervals $[1, 2]$ and $[3, 4]$ are symmetrically placed with respect to $x=2.5$ (as $[2.5-1.5, 2.5-0.5]$ and $[2.5+0.5, 2.5+1.5]$ respectively), the definite integrals over these intervals must be equal.

Verification by direct calculation: First, find the indefinite integral: $\int (x^2 - 5x + 6) dx = \frac{x^3}{3} - \frac{5x^2}{2} + 6x + C$ Let $F(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 6x$.

For the first integral: $\int_{1}^{2} (x^2 - 5x + 6) dx = F(2) - F(1)$ $F(2) = \frac{2^3}{3} - \frac{5(2^2)}{2} + 6(2) = \frac{8}{3} - \frac{20}{2} + 12 = \frac{8}{3} - 10 + 12 = \frac{8}{3} + 2 = \frac{8+6}{3} = \frac{14}{3}$ $F(1) = \frac{1^3}{3} - \frac{5(1^2)}{2} + 6(1) = \frac{1}{3} - \frac{5}{2} + 6 = \frac{2 - 15 + 36}{6} = \frac{23}{6}$ $\int_{1}^{2} (x^2 - 5x + 6) dx = \frac{14}{3} - \frac{23}{6} = \frac{28 - 23}{6} = \frac{5}{6}$

For the second integral: $\int_{3}^{4} (x^2 - 5x + 6) dx = F(4) - F(3)$ $F(4) = \frac{4^3}{3} - \frac{5(4^2)}{2} + 6(4) = \frac{64}{3} - \frac{5(16)}{2} + 24 = \frac{64}{3} - 40 + 24 = \frac{64}{3} - 16 = \frac{64 - 48}{3} = \frac{16}{3}$ $F(3) = \frac{3^3}{3} - \frac{5(3^2)}{2} + 6(3) = \frac{27}{3} - \frac{5(9)}{2} + 18 = 9 - \frac{45}{2} + 18 = 27 - \frac{45}{2} = \frac{54 - 45}{2} = \frac{9}{2}$ $\int_{3}^{4} (x^2 - 5x + 6) dx = \frac{16}{3} - \frac{9}{2} = \frac{32 - 27}{6} = \frac{5}{6}$ Both integrals are equal to $\frac{5}{6}$.

Therefore, the statement is true.

The reason provided in the question is partially correct but could be more precise. The function is symmetric about $x=2.5$, and the integration intervals are symmetric with respect to this axis.