Determine the area enclosed by y=2x+3, the x-axis and the ordinates x=3 and x=4

1 answer:

3 0

Answer:

$\displaystyle \int\limits^4_3 {2x + 3} \, dx = 10$

General Formulas and Concepts:
Calculus

Integration

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Area of a Region Formula: $\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx$

Step-by-step explanation:

Step 1: Define

Identify.

y = 2x + 3

x-interval [3, 4]

x-axis

See attachment for graph.

Step 2: Find Area

Substitute in variables [Area of a Region Formula]: $\displaystyle A = \int\limits^4_3 {2x + 3} \, dx$
[Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle A = \int\limits^4_3 {2x} \, dx + \int\limits^4_3 {3} \, dx$
[Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle A = 2 \int\limits^4_3 {x} \, dx + 3 \int\limits^4_3 {} \, dx$
[Integrals] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle A = 2 \bigg( \frac{x^2}{2} \bigg) \bigg| \limits^4_3 + 3(x) \bigg| \limits^4_3$
[Integrals] Integrate [Integration Rule - FTC 1]: $\displaystyle A = 2 \bigg( \frac{7}{2} \bigg) + 3(1)$
Simplify: $\displaystyle A = 10$