Question

Use the form of the definition of the integral given in the theorem to evaluate the integral. 9 (x2 − 4x + 6) dx

Answer 1

Answer:

$\mathbf{\dfrac{392}{3}}$

Step-by-step explanation:

Given integral :

$\int ^9_1 (x^2 -4x + 6) \ dx$

Let consider $f(x) = x^2 - 4x + 6$ .

Then , using the formula:

$\int ^b_a \ f(x) \ dx = \lim \limits_{x \to \infty} \ \Delta x \ \sum \limits ^{n} _{i=1} \ f(a + i \Delta x)$

where;

a = 1 ,  b = 9 and  $\Delta \ x = \dfrac{b-a}{n}$

$\Delta \ x = \dfrac{9-1}{n}$

$\Delta \ x = \dfrac{8}{n}$

∴

$\int ^9_1 (x^2 -4x + 6) \ dx = \lim \limits _{x \to \infty} \dfrac{8}{n} \ \sum \limits ^{n}_{i=1} \ f(1 + i \dfrac{8}{n})$

= $\lim \limits _{x \to \infty} \dfrac{8}{n} \ \sum \limits ^{n}_{i=1} \ \begin {bmatrix} (1+ i \dfrac{8}{n})^2 - 4(1 + i \dfrac{8}{n} ) +6\end {bmatrix}$

= $\lim \limits _{x \to \infty} \dfrac{8}{n} \ \sum \limits ^{n}_{i=1} \ \begin {bmatrix} 1+ \dfrac{64}{n^2}i^2 + \dfrac{16 }{n} i - 4 - \dfrac{32}{n}i +6\end {bmatrix}$

= $\lim \limits _{x \to \infty} \dfrac{8}{n} \ \sum \limits ^{n}_{i=1} \ \begin {bmatrix} \dfrac{64}{n^2}i^2 - \dfrac{16 }{n} i +3\end {bmatrix}$

= $\lim \limits _{x \to \infty} \dfrac{8}{n} \ \sum \limits ^{n}_{i=1} \ \begin {bmatrix} \dfrac{64}{n^2} \sum \limits ^{n}_{i=1} \ i^2 - \dfrac{16 }{n} \sum \limits ^{n}_{i=1} \ i +3 \sum \limits ^{n}_{i=1} \ 1 \end {bmatrix}$

= $\lim \limits _{x \to \infty} \dfrac{8}{n} \ \sum \limits ^{n}_{i=1} \ \begin {bmatrix} \dfrac{64}{n^2} \dfrac{n(n+1)(2n+1)}{6}-\dfrac{16}{n} \dfrac{n(n+1)}{2} + 3n \end {bmatrix}$

= $\lim \limits _{x \to \infty} \ \begin {bmatrix} \dfrac{512}{6} (1+ \dfrac{1}{n})(2 + \dfrac{1}{n}) -64 (1 + \dfrac{1}{n} ) +24n \end {bmatrix}$

= $\begin {bmatrix} \dfrac{512}{6} (1+ 0)(2 + 0) -64 (1 +0 ) +24 \end {bmatrix}$

= $\begin {bmatrix} \dfrac{512}{6} (2)-64 +24 \end {bmatrix}$

= $\dfrac{512}{3}-40$

= $\dfrac{512- 120}{3}$

= $\mathbf{\dfrac{392}{3}}$