Q 20. - Calculus

Question

When you wish to evaluate the definite integral ${\int }_a^{b}f\left(x\right)dx$ of continuous function $f$, interval $\left[a,b\right]$ is never an impediment to using the Fundamental Theorem of Calculus. However, when you wish to evaluate the double integral ${\iint }_{\Omega }g(x,y)dA$ of a continuous function $g$ over $\Omega$, the region can make the evaluation process easier or harder. Why?

Step-by-Step Solution

Answer

It would be difficult to evaluate functions wrt $x$.

Step 1: Given Information

We need to find difference between definite integral ${\int }_a^{b}f\left(x\right)dx$ and double integral ${\int }_a^{b}f\left(x\right)dx$.

$f$ is continuous function over $\left[a,b\right]$ and $g$ over $\Omega$.

Step 2: Fundamental Theorem of Calculus

Definite integral is calculated using Fundamental Theorem of Calculus.

For calculation of double integral, ${\iint }_{\Omega }g(x,y)dA$, integration is performed wrt $x$ or wrt $y$ according to type of region.

After first integral, it may be easy or hard.

Assume integral is evaluated wrt $x$ first.

The integral along with mentioned limits are

${\iint }_{\Omega }g(x,y)dA={\int }_c^{d{h}_2\left(x\right)}{\int }_{h_1\left(x\right)}g(x,y)dydx$

Solving wrt $y$ first

${\iint }_{\Omega }g(x,y)dA={\int }_c^d\left[G\right(x){]}_{h_2\left(x\right)}^{h_2\left(x\right)}dx$

$={\int }_c^d{\left[G\left(h_2\left(x\right)\right)-G\left(h_1\left(x\right)\right)\right]}_{h_1\left(x\right)}^{h_2\left(x\right)}dx$

and $G\left(x\right)=\int g(x,y)dy$

Functions $G\left(h_2\left(x\right)\right)\text{ and }G\left(h_1\left(x\right)\right)$ can be difficult to evaluate wrt $x$