Answer to Question #100026 in Calculus for Francis

Question

1. Find the area bounded by the curve y= x^2+x+4, the x axis from x=1 to x=3.

Answer 1

The given curve is $y = x^2 + x+ 4$

x is from 1 to 3

Area bounded by the curve and x-axis is A

$$= \int_{x=1}^ {x= 3} y \space dx$$

$$= \int_{x=1}^ {x= 3} (x^2 + x + 4 ) \space dx$$

$$= [ \frac {x^3}{3} + \frac {x^2} {2} + 4x ]_{x=1} ^{x=3}$$

$$= \frac {3^3} {3} + \frac {3^2}{2} + 4(3) - \frac {1}{3} - \frac {1}{2} - 4 (1)$$

$$= 9 + \frac {9}{2} +12 - \frac {1}{3} - \frac {1}{2} - 4$$

$$= 17 + \frac {27-2 -3}{6} = 17 + \frac {11}{3} = \frac {62}{3}$$

Answer:

Area bounded by the curve and x-axis is $\frac {62}{3}$ .