The definite integral $\displaystyle\int_a^b f(x)\,dx$ is a number, not a family of functions — it measures the net signed area between the curve $y=f(x)$ and the $x$-axis from $x=a$ to $x=b$. Area above the axis counts as positive, area below as negative. Here $a$ is the lower limit and $b$ the upper limit.
Definite integral as the limit of a sum
The original definition partitions $[a,b]$ into $n$ equal strips of width $h=\dfrac{b-a}{n}$ and sums the areas of thin rectangles, then lets $n\to\infty$:
$$\int_a^b f(x)\,dx=\lim_{n\to\infty} h\sum_{i=0}^{n-1} f(a+ih),\qquad h=\frac{b-a}{n}.$$
This is why the integral sign $\int$ is an elongated "S" for "sum". Evaluating an integral straight from this definition needs the standard sum formulas $\sum i=\dfrac{n(n-1)}{2}$ and $\sum i^2=\dfrac{(n-1)n(2n-1)}{6}$, but in practice we almost always shortcut it with the theorem below.
Fundamental Theorem of Calculus
If $F$ is any antiderivative of a continuous $f$, then
$$\int_a^b f(x)\,dx=\bigl[F(x)\bigr]_a^b=F(b)-F(a).$$
So: integrate, substitute the upper and lower limits, subtract. The constant $C$ cancels in the subtraction, which is exactly why definite integrals carry no $+C$. This theorem is the bridge between the "area / limit of a sum" idea and the antiderivatives of Topic 1.
Evaluation by substitution — change the limits
When you substitute $u=g(x)$ in a definite integral, you have a choice: either revert to $x$ before applying the limits, or — cleaner — convert the limits too. If $u=g(x)$ then as $x$ runs from $a$ to $b$, $u$ runs from $g(a)$ to $g(b)$:
$$\int_a^b f\bigl(g(x)\bigr)g'(x)\,dx=\int_{g(a)}^{g(b)} f(u)\,du.$$
Once the limits are changed there is no need to substitute back.
Properties of definite integrals
These properties turn many forbidding integrals into one-line evaluations.
For a periodic function with period $T$, $\displaystyle\int_0^{nT} f(x)\,dx=n\int_0^{T} f(x)\,dx$ and $\displaystyle\int_a^{a+T} f(x)\,dx=\int_0^{T} f(x)\,dx$ for every $a$. The King's property (P4) is the workhorse: replacing $x$ by $a-x$ often produces a second copy of the integral, which you add to the original to solve for it.
Deeper Insight — the properties are symmetries, not tricks to memorise. Each property says something geometric. P1 is "walking the interval backwards flips the sign of area". P7 is the visible symmetry of even and odd graphs about the $y$-axis — an odd function's left half exactly cancels its right half, so the integral is zero before you compute anything. The King's property P4 is a reflection of the strip $[0,a]$ about its midpoint, which leaves the total area unchanged but swaps the integrand for $f(a-x)$. Recognising these symmetries first can save an entire page of algebra: the smartest move on a hard definite integral is usually to test for odd/even symmetry or apply King's property before reaching for an antiderivative.
Evaluate $\displaystyle\int_0^2 x^2\,dx$.
SolutionAntiderivative $\dfrac{x^3}{3}$. So $\left[\dfrac{x^3}{3}\right]_0^2=\dfrac{8}{3}-0=\dfrac{8}{3}.$
Answer: $\dfrac{8}{3}$
Evaluate $\displaystyle\int_0^{\pi} \sin x\,dx$.
Solution$[-\cos x]_0^{\pi}=-\cos\pi-(-\cos0)=-(-1)+1=2.$
Answer: $2$
Evaluate $\displaystyle\int_{-1}^{1} x^3\,dx$.
Solution$x^3$ is an odd function and the limits are symmetric, so by property P7 the integral is $0$.
Answer: $0$
Evaluate $\displaystyle\int_1^3 (2x+1)\,dx$.
SolutionAntiderivative $x^2+x$. $[x^2+x]_1^3=(9+3)-(1+1)=12-2=10.$
Answer: $10$
Evaluate $\displaystyle\int_1^4 \frac{1}{\sqrt{x}}\,dx$.
Solution$\dfrac{1}{\sqrt{x}}=x^{-1/2}$ has antiderivative $2\sqrt{x}$. So $[2\sqrt{x}]_1^4=2(2)-2(1)=4-2=2.$
Answer: $2$
Evaluate $\displaystyle\int_1^e \frac{1}{x}\,dx$.
Solution$[\ln|x|]_1^e=\ln e-\ln 1=1-0=1.$
Answer: $1$
Evaluate $\displaystyle\int_0^{\pi/4} \sec^2 x\,dx$.
SolutionAntiderivative $\tan x$. $[\tan x]_0^{\pi/4}=\tan\dfrac{\pi}{4}-\tan 0=1-0=1.$
Answer: $1$
Evaluate $\displaystyle\int_0^1 x\,dx$ as the limit of a sum.
SolutionHere $a=0,\ b=1,\ h=\dfrac1n,\ f(x)=x$. So $\displaystyle\int_0^1 x\,dx=\lim_{n\to\infty}\frac1n\sum_{i=0}^{n-1}\frac{i}{n}=\lim_{n\to\infty}\frac{1}{n^2}\cdot\frac{n(n-1)}{2}=\lim_{n\to\infty}\frac{n-1}{2n}=\frac12.$ (The FTC confirms: $[\tfrac{x^2}{2}]_0^1=\tfrac12$.)
Answer: $\dfrac12$
Evaluate $\displaystyle\int_0^{\pi/2} \frac{\cos x}{1+\sin x}\,dx$.
SolutionPut $u=1+\sin x$, $du=\cos x\,dx$. Limits: $x=0\Rightarrow u=1$; $x=\tfrac{\pi}{2}\Rightarrow u=2$. So $\displaystyle\int_1^2\frac{du}{u}=[\ln u]_1^2=\ln 2.$
Answer: $\ln 2$
Evaluate $\displaystyle\int_0^1 x\,e^{x^2}\,dx$.
SolutionPut $u=x^2$, $du=2x\,dx$, so $x\,dx=\tfrac12\,du$. Limits: $x=0\Rightarrow u=0$; $x=1\Rightarrow u=1$. $\displaystyle\frac12\int_0^1 e^{u}\,du=\frac12[e^u]_0^1=\frac12(e-1).$
Answer: $\dfrac12(e-1)$
Evaluate $\displaystyle\int_0^{\pi/4} \tan^2 x\,dx$.
SolutionUse $\tan^2x=\sec^2x-1$. $\displaystyle\int_0^{\pi/4}(\sec^2x-1)\,dx=[\tan x-x]_0^{\pi/4}=\left(1-\frac{\pi}{4}\right)-0=1-\frac{\pi}{4}.$
Answer: $1-\dfrac{\pi}{4}$
Evaluate $\displaystyle\int_0^{\pi/2} \frac{\sin x}{\sin x+\cos x}\,dx$.
SolutionCall it $I$. By King's property ($x\to\tfrac{\pi}{2}-x$), $\displaystyle I=\int_0^{\pi/2}\frac{\cos x}{\cos x+\sin x}\,dx$. Adding the two forms, $\displaystyle 2I=\int_0^{\pi/2}\frac{\sin x+\cos x}{\sin x+\cos x}\,dx=\int_0^{\pi/2}1\,dx=\frac{\pi}{2}$, so $I=\dfrac{\pi}{4}.$
Answer: $\dfrac{\pi}{4}$
Evaluate $\displaystyle\int_0^{\pi} \frac{x\,\sin x}{1+\cos^2 x}\,dx$.
SolutionBy King's property ($x\to\pi-x$), $\displaystyle I=\int_0^{\pi}\frac{(\pi-x)\sin x}{1+\cos^2 x}\,dx$. Adding, $\displaystyle 2I=\pi\int_0^{\pi}\frac{\sin x}{1+\cos^2 x}\,dx$. Put $t=\cos x$: $\displaystyle 2I=\pi\int_{-1}^{1}\frac{dt}{1+t^2}=\pi[\tan^{-1}t]_{-1}^{1}=\pi\cdot\frac{\pi}{2}$, so $I=\dfrac{\pi^2}{4}.$
Answer: $\dfrac{\pi^2}{4}$
Evaluate $\displaystyle\int_{-\pi/2}^{\pi/2} \sin^3 x\cos^2 x\,dx$.
Solution$\sin^3x\cos^2x$ is odd ($\sin$ is odd, $\cos$ even, odd$\times$even$=$odd) and the limits are symmetric, so by P7 the integral is $0$ — no calculation needed.
Answer: $0$
Evaluate $\displaystyle\int_{-2}^{2} (x^4+3x^2)\,dx$.
Solution$x^4+3x^2$ is even, so by P7 $\displaystyle\int_{-2}^{2}=2\int_0^{2}(x^4+3x^2)\,dx=2\left[\frac{x^5}{5}+x^3\right]_0^2=2\left(\frac{32}{5}+8\right)=\frac{144}{5}.$
Answer: $\dfrac{144}{5}$
Evaluate $\displaystyle\int_0^{2\pi} \cos x\,dx$ and $\displaystyle\int_0^{2\pi}\sin^2 x\,dx$.
Solution$[\sin x]_0^{2\pi}=0$. For the second, $\sin^2x=\dfrac{1-\cos2x}{2}$, so $\displaystyle\int_0^{2\pi}\frac{1-\cos2x}{2}\,dx=\frac12\left[x-\frac{\sin2x}{2}\right]_0^{2\pi}=\frac12(2\pi)=\pi.$
Answer: $\displaystyle\int_0^{2\pi}\cos x\,dx=0$ and $\displaystyle\int_0^{2\pi}\sin^2 x\,dx=\pi$
Evaluate $\displaystyle\int_0^{\pi} x\,\sin x\,dx$.
SolutionBy parts $u=x,\ dv=\sin x\,dx$: $v=-\cos x$. $\displaystyle\int_0^{\pi}x\sin x\,dx=[-x\cos x]_0^{\pi}+\int_0^{\pi}\cos x\,dx=(\pi)+[\sin x]_0^{\pi}=\pi+0=\pi.$
Answer: $\pi$
Evaluate $\displaystyle\int_0^{\pi/2} \sin^2 x\,dx$.
Solution$\sin^2x=\dfrac{1-\cos2x}{2}$. $\displaystyle\int_0^{\pi/2}\frac{1-\cos2x}{2}\,dx=\frac12\left[x-\frac{\sin2x}{2}\right]_0^{\pi/2}=\frac12\left(\frac{\pi}{2}-0\right)=\frac{\pi}{4}.$
Answer: $\dfrac{\pi}{4}$
Evaluate $\displaystyle\int_2^3 \frac{dx}{x^2-1}$.
SolutionUsing $\int\dfrac{dx}{x^2-1}=\dfrac12\ln\left|\dfrac{x-1}{x+1}\right|$, $\displaystyle\int_2^3=\frac12\left[\ln\frac{x-1}{x+1}\right]_2^3=\frac12\left(\ln\frac{2}{4}-\ln\frac{1}{3}\right)=\frac12\ln\frac{3}{2}.$
Answer: $\dfrac12\ln\dfrac{3}{2}$
Evaluate $\displaystyle\int_0^{\pi} \frac{x}{1+\sin x}\,dx$.
SolutionBy King's property ($x\to\pi-x$, using $\sin(\pi-x)=\sin x$), $\displaystyle I=\int_0^{\pi}\frac{\pi-x}{1+\sin x}\,dx$. Adding, $\displaystyle 2I=\pi\int_0^{\pi}\frac{dx}{1+\sin x}$. Multiply numerator and denominator by $(1-\sin x)$: $\dfrac{1-\sin x}{\cos^2x}=\sec^2x-\sec x\tan x$, whose antiderivative is $\tan x-\sec x$. Then $\displaystyle\int_0^{\pi}\frac{dx}{1+\sin x}=[\tan x-\sec x]_0^{\pi}=2$, so $2I=2\pi$ and $I=\pi.$
Answer: $\pi$