Question

evaluate the following integral using the fundamental theorem of calculus. explain why the result is consistent with the figure.

1. (int_{0}^{1}(x^{2}-3x + 6)dx)

(int_{0}^{1}(x^{2}-3x + 6)dx=square) (type an exact answer.)
consider the trapezoid whose bases are the left and right sides of the shaded region. the trapezoid has a longer base of length (a=square), a shorter base of length (b=square), and a height of (h=square) so the area of the trapezoid is (square). the value of the integral found in the previous step is slightly than the area of the trapezoid, as the figure suggests it should be
(simplify your answers.)

Explanation:

Step1: Find antiderivative

The integral is $\int_{0}^{1}(x^2 - 3x + 6)dx$. The antiderivative of $x^2$ is $\frac{x^3}{3}$, of $-3x$ is $-\frac{3x^2}{2}$, and of $6$ is $6x$. So the antiderivative $F(x)=\frac{x^3}{3}-\frac{3x^2}{2}+6x$.

Step2: Apply Fundamental Theorem

Evaluate $F(1)-F(0)$.
$F(1)=\frac{1^3}{3}-\frac{3(1)^2}{2}+6(1)=\frac{1}{3}-\frac{3}{2}+6$.
$F(0)=\frac{0^3}{3}-\frac{3(0)^2}{2}+6(0)=0$.
Calculate $F(1)$: $\frac{1}{3}-\frac{3}{2}+6=\frac{2 - 9 + 36}{6}=\frac{29}{6}\approx4.833$.

Step3: Analyze trapezoid

For the trapezoid, left base (at $x = 0$): $y(0)=0^2 - 3(0)+6 = 6$.
Right base (at $x = 1$): $y(1)=1^2 - 3(1)+6 = 4$.
Height $h = 1 - 0 = 1$.
Area of trapezoid: $\frac{(a + b)}{2}h=\frac{(6 + 4)}{2}(1)=5$.
The integral value $\frac{29}{6}\approx4.833$ is slightly less than $5$ because the curve is concave up (second derivative $y'' = 2>0$), so the trapezoid overestimates the area under the curve.

Answer:

$\int_{0}^{1}(x^2 - 3x + 6)dx=\frac{29}{6}$ (or approximately $4.83$). The trapezoid has $a = 6$, $b = 4$, $h = 1$, area $5$. The integral value is slightly less than the trapezoid area.