Answer to Question #129613 in Calculus for Jade

Question

Find the area between the x-axis and the graph of f(x)=x^3-1 from x= 0 to x=4

Answer 1

As per the question,

Let y=f(x)=x$^3$-1

at x=0,y=-1

also

at y=0,x=1

The given curve passes through(0,-1),(1,0)

Therefore the area under these curve is divided into two parts

i.e. from x=0 to x=1 and from x=1 to x=4

Area= $\mid$ $\intop_{0}^{1}$ f(x)dx$\mid$ +$\intop_{1}^{4}$ f(x)dx

=$\mid\intop_{0}^{1}$ (x$^3$ -1)dx$\mid$ + $\intop_{1}^{4}$ (x$^3-$ 1)dx

= $\mid$ $\frac{x^4}{4}-x\mid_{0}^{1} + \mid\frac{x^4}{4}-x\mid_1^{4}$

Putting the upper and lower limits we get

=$\mid\frac{1}{4}-1\mid$ +$(\frac{256}{4}-4)-(\frac{1}{4}-1)$

=$\frac{3}{4}+\frac{3}{4}+60$

=0.75+0.75+60

=61.5 square units

Hence the area under curve is 61.5 sq. units