Using fundamental Theorem of calculus to evaluate $$\int^{15}_{6}(6-x)(x-15)dx$$
What i try:
$$I =\int^{15}_{6}(6-x)(x-15)dx$$
$$I =\int^{15}_{6}(21x-x^2-90)dx=\bigg(\frac{21}{2}x^2-\frac{x^3}{3}-90x\bigg)\bigg|^{15}_{6}$$
I did not understand why this method is wrong. What is the Right method to evaluate that integration.plesee explain me. Thank
Fundamental theorem of calculus:
First part: let $f$ be continuous on $[a,b]$. If $F$ is defined on $[a,b]$ as $$F(x)=\int _a^x f(t)dt$$, then $F$ is uniformly continuous and differentiable on $[a,b]$, and for all $x^*\in[a,b]$, $$\frac {dF} {dx} \vert _{x=x^*} = f(x^*)$$.
Second part: If $$\frac {dF} {dx} \vert _{x=x^*}$$ for all $x^*\in[a,b]$, and $f$ is Riemann integrable, then $$\int _a ^b f(x)dx = F(b)-F(a)$$.
Conclusion: your answer is good...
