Question

\int_{1}^{4} \frac{x + 1}{\sqrt{x}} dx

Explanation:

Step1: Simplify the integrand

First, we simplify the integrand \(\frac{x + 1}{\sqrt{x}}\). We can split the fraction:

$$ \frac{x + 1}{\sqrt{x}}=\frac{x}{\sqrt{x}}+\frac{1}{\sqrt{x}} $$

Simplify each term:

$$ \frac{x}{\sqrt{x}}=x^{1-\frac{1}{2}} = x^{\frac{1}{2}}=\sqrt{x} $$

$$ \frac{1}{\sqrt{x}}=x^{-\frac{1}{2}} $$

So the integrand becomes \(x^{\frac{1}{2}}+x^{-\frac{1}{2}}\).

Step2: Integrate term by term

Now we integrate the simplified integrand with respect to \(x\) from \(1\) to \(4\). The integral of \(x^n\) is \(\frac{x^{n + 1}}{n+1}+C\) (we can ignore the constant \(C\) for definite integrals).

For the first term \(x^{\frac{1}{2}}\):

$$ \int x^{\frac{1}{2}}dx=\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}=\frac{2}{3}x^{\frac{3}{2}} $$

For the second term \(x^{-\frac{1}{2}}\):

$$ \int x^{-\frac{1}{2}}dx=\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=\frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 2x^{\frac{1}{2}} $$

So the antiderivative \(F(x)\) of the integrand is:

$$ F(x)=\frac{2}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}} $$

Step3: Evaluate the definite integral

We use the Fundamental Theorem of Calculus, which states that \(\int_{a}^{b}f(x)dx=F(b)-F(a)\).

First, we find \(F(4)\):

$$ F(4)=\frac{2}{3}(4)^{\frac{3}{2}}+2(4)^{\frac{1}{2}} $$

Calculate \(4^{\frac{3}{2}}=(4^{\frac{1}{2}})^3 = 2^3 = 8\) and \(4^{\frac{1}{2}} = 2\). So:

$$ F(4)=\frac{2}{3}\times8+2\times2=\frac{16}{3}+4=\frac{16 + 12}{3}=\frac{28}{3} $$

Next, we find \(F(1)\):

$$ F(1)=\frac{2}{3}(1)^{\frac{3}{2}}+2(1)^{\frac{1}{2}} $$

Since \(1^{\frac{3}{2}} = 1\) and \(1^{\frac{1}{2}} = 1\), we have:

$$ F(1)=\frac{2}{3}\times1+2\times1=\frac{2}{3}+2=\frac{2 + 6}{3}=\frac{8}{3} $$

Now, we subtract \(F(1)\) from \(F(4)\):

$$ \int_{1}^{4}\frac{x + 1}{\sqrt{x}}dx=F(4)-F(1)=\frac{28}{3}-\frac{8}{3}=\frac{20}{3} $$

Answer:

\(\frac{20}{3}\)