Question

evaluate the definite integral. use a graphing utility to verify your result.
$$ int _ { 0 } ^ { 1 } \frac { x - sqrt { x } } { 7 } d x $$

Explanation:

Step1: Simplify the integrand

We can split the fraction: $\frac{x - \sqrt{x}}{7}=\frac{1}{7}x-\frac{1}{7}x^{\frac{1}{2}}$

Step2: Integrate term by term

The integral of $\frac{1}{7}x$ with respect to $x$ is $\frac{1}{7}\cdot\frac{x^{2}}{2}=\frac{x^{2}}{14}$, and the integral of $\frac{1}{7}x^{\frac{1}{2}}$ with respect to $x$ is $\frac{1}{7}\cdot\frac{x^{\frac{3}{2}}}{\frac{3}{2}}=\frac{2x^{\frac{3}{2}}}{21}$
So the antiderivative $F(x)$ is $\frac{x^{2}}{14}-\frac{2x^{\frac{3}{2}}}{21}$

Step3: Apply the Fundamental Theorem of Calculus

Evaluate $F(1)-F(0)$
$F(1)=\frac{1^{2}}{14}-\frac{2\cdot1^{\frac{3}{2}}}{21}=\frac{1}{14}-\frac{2}{21}$
$F(0)=\frac{0^{2}}{14}-\frac{2\cdot0^{\frac{3}{2}}}{21}=0 - 0=0$
Now, find a common denominator for $\frac{1}{14}-\frac{2}{21}$, which is 42.
$\frac{1}{14}=\frac{3}{42}$ and $\frac{2}{21}=\frac{4}{42}$
So $\frac{3}{42}-\frac{4}{42}=-\frac{1}{42}$

Answer:

$-\frac{1}{42}$