\[5.\;\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx\]
\[5.\;\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx\]
Given a continuous function \(f\) on \([a,b]\), define a new function \[g(x)=\int_{a}^x f(t)dt\] where \(x\) varies between \(a\) and \(b\). Then \(g(x)\) is the (signed) area under the curve \(f\) over \([a,x]\).
The Fundamental Theorem of Calculus(FTC) - Part 1
If \(f\) is continuous on \([a,b]\) then the function \(g\) defined above is continuous on \([a,b]\) and is differentiable on \((a,b)\), and
\[g'(x)=f(x).\]
Example 1.
Find the derivative of the function \(g(x)=\int_{-10}^x \ln(1+t^2) dt\).
Sol: \(\ln(1+x^2)\) is continuous, so FTC-1 implies that \(g'(x)=\ln(1+x^2)\).
Example 2.
Find the derivative of the function
\(G(x)=\int_{-10}^{x^2} \ln(1+t^2) dt\).
Sol: We cannot repeat the argument in Example 1 because the upper limit is \(x^2\) instead of \(x\). Chain rule has to be used here.
Let \(u=x^2\). Then \(G(x)=g(u)=\int_{-1}^u \ln(1+t^2)dt\) and \(\dfrac{dg}{du}=\ln(1+u^2).\)
\[\dfrac{dG}{dx}=\dfrac{dg}{du}\cdot \dfrac{du}{dx}=\ln(1+u^2)\cdot (2x) \]
The Fundamental Theorem of Calculus(FTC) - Part 2
If \(f\) is continuous on \([a,b]\), then
\[\int_a^b f(x)dx=F(x)|_a^b = F(b)-F(a)\]
where \(F\) is any antiderivative of \(f\), that is, \(F\) is a function such that \(F'=f\).
Example 1.
Evaluate the integral \(\int_2^{10} \frac{1}{x} dx\).
Sol: An antiderivative of \(\frac{1}{x}\) is \(\ln |x|\). FTC-2 tells us that \[\int_2^{10} \frac{1}{x} dx=\ln 10 - \ln 2 =\ln 5\]
Example 2.
Find the area under the curve \(y=\sqrt{x}\) from \(0\) to \(1\).
Sol: An antiderivative of \(y=\sqrt{x}=x^{1/2}\) is \(\frac{2}{3}x^{3/2}\). The area is equal to
\[\int_0^1\sqrt{x}dx = \frac{2}{3}x^{3/2}|_0^1=\frac{2}{3}-0=\frac{2}{3}\]
Example 3.
The velocity of a snail is \(v(t)=\sqrt{t}\).
Find the distance it traveled from \(t=0\) to \(t=1\).
Sol: The distance traveled is
\[\int_0^1 v(t)dt=\int_0^1\sqrt{t}dt=\frac{2}{3},\]
same integral as in Example 2..