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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

User KaoD
by
6.7k points

1 Answer

3 votes

Answer:

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

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 = (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

=
((x^(6) )/(6) + (8 x^(5) )/(5) + 2 (x^(3) )/(3) +(5 x^(2) )/(2) + 15 x)^(6) _(-6)


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

=
((6^(6) )/(6) + (8 6^(5) )/(5) + 2 (6^(3) )/(3) +(5 6^(2) )/(2) + 15X 6) - ((((-6)^(6) )/(6) + (8 (-6)^(5) )/(5) + 2 ((-6)^(3) )/(3) +(5 (-6)^(2) )/(2) + 15 (-6))

After simplification and cancellation we get

=
(2 X 8 X (6)^(5) )/(5) + (2 X 2 X (6)^3)/(3) + 2 X 15 X 6

on calculation , we get

=
(124,416)/(5) + (864)/(3) + 180

On L.C.M 15

=
(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

User Mateusz Nowak
by
6.4k points
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