Calculus Application: Deriving the Area of a Triangle through Definite Integration

APPLICATION OF DEFINITE INTEGRAL

• Definite integration serves as a foundational tool in calculus for calculating various geometric and physical quantities, such as areas, volumes, work, and center of mass.
• One principal application is determining the total area of a geometric figure by integrating the lengths of infinitesimal cross-sectional strips.

AREA OF A TRIANGLE

• The objective is to derive the standard geometric formula for the area of a triangle using the principles of integration.
• The known geometric formula for the area ($A$) is:   - $A = \frac{1}{2} b h_0$   - Where $b$ is the base of the triangle.   - Where $h_0$ is the vertical height of the triangle.

SETTING UP THE INTEGRAL

• To calculate the area through integration, we consider a triangle with height $h_0$ and base $b$.
• We define a horizontal slice (strip) at an arbitrary height $h$ from the vertex.
• Let the length of this strip be $L$. By the property of similar triangles, the relationship between the strip length $L$ and the base $b$ relative to the height is expressed as:   - $L = b \left( \frac{h}{h_0} \right)$
• The infinitesimal area ($dA$) of this strip is given by the product of its length ($L$) and its infinitesimal height ($dh$):   - $dA = L \, dh = \left( \frac{b}{h_0} h \right) dh$

INTEGRATION STEPS AND EVALUATION

• The total area ($A$) is obtained by integrating the strip area ($dA$) from the vertex ($h = 0$) to the base ($h = h_0$):   - $A = \int_0^{h_0} \frac{b h}{h_0} dh$
• The term $\frac{b}{h_0}$ is a constant with respect to the variable of integration $h$, and can thus be factored out of the integral:   - $A = \frac{b}{h_0} \int_0^{h_0} h \, dh$
• Applying the power rule for integration, where $\int h^n \, dh = \frac{h^{n+1}}{n+1}$:   - Here, $n = 1$, so the integral becomes:     - $A = \frac{b}{h_0} \left[ \frac{h^{1+1}}{1+1} \right]_0^{h_0}$     - $A = \frac{b}{h_0} \left[ \frac{h^2}{2} \right]_0^{h_0}$
• We evaluate the definite integral by substituting the upper limit ($h_0$) and the lower limit ($0$):   - $A = \frac{b}{h_0} \left( \frac{h_0^2}{2} - \frac{0^2}{2} \right)$   - $A = \frac{b}{h_0} \left( \frac{h_0^2}{2} \right)$
• Simplifying the resulting expression:   - $A = \frac{b \times h_0^2}{h_0 \times 2}$   - One factor of $h_0$ in the numerator cancels with the $h_0$ in the denominator:     - $A = \frac{b h_0}{2}$     - $A = \frac{1}{2} b h_0$
• The derivation concludes that the area found via definite integration matches the classical geometric formula exactly.