/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 33

Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{1}^{9} \frac{2}{\sqrt{x}} d x$$

Short Answer

Expert verified
Answer: 4鈭歺 + C

Step by step solution

01

Finding the antiderivative of the integrand function

To find the antiderivative of \(\frac{2}{\sqrt{x}}\), we can first rewrite the function as \(2x^{-\frac{1}{2}}\). Now, we can use the power rule for antiderivatives: $$\int x^n dx = \frac{x^{n+1}}{n+1} + C$$ Applying the power rule to our function gives: $$\int 2x^{-\frac{1}{2}} dx = 2\int x^{-\frac{1}{2}} dx = 2\left(\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right) + C = 4\sqrt{x} + C$$
02

Applying the Fundamental Theorem of Calculus

Now that we have the antiderivative of the integrand function, we can apply the Fundamental Theorem of Calculus. It states that if F'(x) = f(x), then: $$\int_a^b f(x) dx = F(b) - F(a)$$ We have determined that the antiderivative of the integrand function, F(x), is \(4\sqrt{x} + C\). So, we can find the value of the integral as follows: $$\int_{1}^{9} \frac{2}{\sqrt{x}} d x = \left[4\sqrt{x}\right]_1^9$$
03

Evaluate the integral

Now we can substitute the values of the limits into the antiderivative we found in step 1 to evaluate the integral: $$\left[4\sqrt{x}\right]_1^9 = 4\sqrt{9} - 4\sqrt{1} = 4(3) - 4(1) = 12 - 4 = 8$$ Thus, the value of the given integral is 8.

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