(1 point) find the exact area of the region that lies beneath the curve y = 5√x, 0 ≤ x ≤ 27.
(1 point) find the exact area of the region that lies beneath the curve y = 5√x, 0 ≤ x ≤ 27.
Answer
Explanation:
Step1: Recall the area - under - curve formula
The area (A) under the curve (y = f(x)) from (x=a) to (x = b) is given by (A=\int_{a}^{b}f(x)dx). Here, (f(x)=5\sqrt{x}=5x^{\frac{1}{2}}), (a = 0), and (b = 27).
Step2: Use the power - rule for integration
The power - rule for integration is (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C) ((n\neq - 1)). So, (\int5x^{\frac{1}{2}}dx=5\times\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C=\frac{10}{3}x^{\frac{3}{2}}+C).
Step3: Evaluate the definite integral
(\int_{0}^{27}5x^{\frac{1}{2}}dx=\left[\frac{10}{3}x^{\frac{3}{2}}\right]_{0}^{27}). Substitute the upper and lower limits: (\frac{10}{3}(27)^{\frac{3}{2}}-\frac{10}{3}(0)^{\frac{3}{2}}). Since ((27)^{\frac{3}{2}}=(3^{3})^{\frac{3}{2}}=3^{\frac{9}{2}} = 3^{4}\times3^{\frac{1}{2}}=81\sqrt{3}), then (\frac{10}{3}(27)^{\frac{3}{2}}=\frac{10}{3}\times81\sqrt{3}=270\sqrt{3}).
Answer:
(270\sqrt{3})