Question

Use the Fundamental Theorem of Calculus to find the area of the region between the graph of the function x5 + 8x4 + 2x2 + 5x + 15 and the x-axis on the interval [–6, 6]. Round off your answer to the nearest integer.
Question 15 options:

25,351 units2

149,473 units2

3,758 units2

2,362 units2

Answer 1

Answer:

The area of the region between the graph of the given function and the x-axis = 25,351 units²

Step-by-step explanation:

Given  x⁵ + 8 x⁴ + 2 x² + 5 x + 15

If 'f' is a continuous on [a ,b] then the function

            $F(x) = \int\limits^a_b {f(x)} \, dx$

By using integration formula

$\int{x^n} \, dx = \frac{x^{n+1} }{n+1} +c$

Given  x⁵ + 8 x⁴ + 2 x² + 5 x + 15 in the interval [-6,6]

 $\int\limits^6_^-6} (x^{5} + 8 x^{4} + 2 x^{2} + 5 x + 15) )dx$

On integration , we get

=   $(\frac{x^{6} }{6} + \frac{8 x^{5} }{5} + 2 \frac{x^{3} }{3} +\frac{5 x^{2} }{2} + 15 x)^{6} _{-6}$

$F(x) = \int\limits^a_b {f(x)} \, dx = F(b) -F(a)$

= $(\frac{6^{6} }{6} + \frac{8 6^{5} }{5} + 2 \frac{6^{3} }{3} +\frac{5 6^{2} }{2} + 15X 6) - ((\frac{(-6)^{6} }{6} + \frac{8 (-6)^{5} }{5} + 2 \frac{(-6)^{3} }{3} +\frac{5 (-6)^{2} }{2} + 15 (-6))$

After simplification and cancellation we get

 =  $\frac{2 X 8 X (6)^{5} }{5} + \frac{2 X 2 X (6)^3}{3} + 2 X 15 X 6$

on calculation , we get

= $\frac{124,416}{5} + \frac{864}{3} + 180$

On L.C.M  15

= $\frac{124,416 X 3 + 864 X 5 + 180 X 15}{15}$

= 25 351.2 units²

Conclusion:-

The area of the region between the graph of the given function and the x-axis = 25,351 units²