How can a definite integral be negative? \[\]

Hint: We recall the definition of definite integral of function $f\left( x \right)$ with respect to $x$ within the interval $\left[ a,b \right]$ as the area bounded by the curve and the lines $x=a,x=b$. We recall that area will obtained as negative if all parts of bounded region lie below $x-$axis or more parts of the region lie below $x-$axis then above $x-$axis.\[\]

Complete step-by-step answer:
We know that integral or primitive function of $f\left( x \right)$ is given as $F\left( x \right)+c$ where$\dfrac{d}{dx}F\left( x \right)=f\left( x \right)$. If we integrate within a certain interval $x\in \left[ a,b \right]$ rather than all over the domain then we call it a definite integral and we express it as
\[\int_{a}^{b}{f\left( x \right)}=\left[ F\left( x \right) \right]_{a}^{b}=F\left( b \right)-F\left( a \right)\]
The definite integral of the function $f\left( x \right)$ is the area of the enclosed region by the curve within the area bounds$x=a,x=b$. If all of the enclosed region lie above the $x-$axis then area as well as definite integral will be positive which means
\[\int_{a}^{b}{f\left( x \right)}\ge 0\text{ if }f\left( x \right)\ge 0\text{ for }x\in \left[ a,b \right]\]
We can take an example $f\left( x \right)=\sqrt{x}$ which is positive for the defined domain $x\in \left( 0,\infty \right)$. \[\]
 

If all of the enclosed region lies blow the $x-$axis then area as well as definite integral will be negative which means
\[\int_{a}^{b}{f\left( x \right)}\le 0\text{ if }f\left( x \right)\le 0\text{ for }x\in \left[ a,b \right]\]
We can take an example $f\left( x \right)=-\sqrt{x}$ which is negative for the defined domain $x\in \left( 0,\infty \right)$. \[\]
 

If the area of the enclosed region that lies below the $x-$axis is more than the area above the $x-$axis then the definite integral will be negative. If the curve intersects at some $x=c\in \left[ a,b \right]$ then we assume the area above as ${{A}_{a}}=\int_{a}^{c}{f\left( x \right)}$ and area below ${{A}_{b}}=\int_{c}^{b}{f\left( x \right)}$. Then we have
\[\int_{a}^{b}{f\left( x \right)}=\int_{a}^{c}{f\left( x \right)}+\int_{c}^{b}{f\left( x \right)}={{A}_{a}}+{{A}_{b}}\le 0\text{ if }\left| {{A}_{a}} \right|\le \left| {{A}_{b}} \right|\]
 Let us consider $f\left( x \right)=\sin x,a=-\pi ,b=\dfrac{-\pi }{2}$ as an example. We can represent $\int_{-\pi }^{-\dfrac{-\pi }{2}}{f\left( x \right)}<0$ as the area shaded below. \[\]



Note: We note that the definite integration of $x=f\left( y \right)$within interval $y\in \left[ a,b \right]$with respect to $y$ will be negative if the enclosed region by the curve $f\left( y \right)$ and the bounds $y=a,y=b$ will be at the left side of $y-$axis or the area at the left side will be more than area at the right side. We can find a definite integral of functions whose indefinite integral cannot be determined with approximation.