question 5\n2 pts\nwhich of the following is the area under the curve y = \\frac{1}{\\sqrt{x}} between x = 1…

question 5\n2 pts\nwhich of the following is the area under the curve y = \\frac{1}{\\sqrt{x}} between x = 1 and x = 2?\nnone of these\n\\ln\\sqrt{2}\n\\frac{\\sqrt{2}-1}{2}\n\\frac{\\ln\\sqrt{2}}{\\sqrt{2}}\n2\\sqrt{2}-2

question 5\n2 pts\nwhich of the following is the area under the curve y = \\frac{1}{\\sqrt{x}} between x = 1 and x = 2?\nnone of these\n\\ln\\sqrt{2}\n\\frac{\\sqrt{2}-1}{2}\n\\frac{\\ln\\sqrt{2}}{\\sqrt{2}}\n2\\sqrt{2}-2

Answer

Explanation:

Step1: Recall 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, $f(x)=\frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}$, $a = 1$, and $b = 2$.

Step2: Calculate the definite integral

We know that $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ for $n\neq - 1$. So, $\int_{1}^{2}x^{-\frac{1}{2}}dx=\left[2x^{\frac{1}{2}}\right]_{1}^{2}$.

Step3: Evaluate the definite - integral

Substitute the upper and lower limits: $2\sqrt{2}-2\sqrt{1}=2\sqrt{2}-2$.

Answer:

$2\sqrt{2}-2$