Question

How to find area of curve using definite integral?

Answer 1

The drawn curve here represents $f\left(x\right)=x^2+3x+4$


Now, let us say, you want area covered by this curve and x-axis between x=1 and x=3, it is shown ln hatched portion.
$F\left(x\right)=x^2+3x+4$
so we integrate it, between x=1 and x=3
${\int }_1^3(s^2+3x+4)dx=\frac{x^2}{3}+3\frac{x^2}{2}+4.x^2$
Now the inter and on R.H.S for x=3 first and then subtract the value of R.H.S for x=3. This gives you the area between the curve fof $x^2+3x+4$ and x-axis between x=1 and x=3.
$\to (\frac{3^3}{3}+\frac{3^2}{2+4,3})-(\frac{1}{3}+3.\frac{1}{2}+4)\to (9+13.5+12)-(0.333+1.5+4)\to 34.5-5.833=28.667sq.units$
Area under curve $x^2+3x+4$ between x=1 and $x=3-{\int }_1^3(s^2+3x+4)dx$
=28.667 sq. units (as calculated above)