Question

How can a definite integral be negative? $$$$

Answer 1

Hint: We recall the definition of definite integral of function $f\left(x\right)$ with respect to $x$ within the interval $[a,b]$ 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${\displaystyle \frac{d}{dx}}F\left(x\right)=f\left(x\right)$. If we integrate within a certain interval $x\in [a,b]$ 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
$${\int }_a^{b}f\left(x\right)\ge 0\text{ if }f\left(x\right)\ge 0\text{ for }x\in [a,b]$$
We can take an example $f\left(x\right)=\sqrt{x}$ which is positive for the defined domain $x\in (0,\mathrm{\infty })$. $$$$
 

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

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 $x=c\in [a,b]$ 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 $f\left(x\right)=\sin x,a=-\pi ,b={\displaystyle \frac{-\pi }{2}}$ as an example. We can represent ${\int }_{-\pi }^{-{\displaystyle \frac{-\pi }{2}}}f\left(x\right)<0$ as the area shaded below. $$$$

Note: We note that the definite integration of $x=f\left(y\right)$within interval $y\in [a,b]$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.