Fundamental Theorem of Integral Calculus

The Fundamental Theorem of Integral Calculus connects differentiation and integration. It shows that a definite integral can be evaluated by finding an antiderivative of the given function and using the values at the upper and lower limits.

It turns area-based integral ideas into a quick calculation method used in board and entrance examinations.

If a function f is continuous on an interval, the area function is defined by

This means that A(x) gives the area accumulated from x = a to a variable point x.

If f is continuous on [a, b] and

If f is continuous on [a, b] and F is any antiderivative of f, then

This is the formula most often used in exams to evaluate definite integrals.

\[\int_{4}^{9} \frac{\sqrt{x}}{(30 - x^{\frac{3}{2}})^2} dx\]

Solution:

Let \[I = \int_{4}^{9} \frac{\sqrt{x}}{(30 - x^{\frac{3}{2}})^2} dx\].

We first find the anti-derivative of the integrand.

Put \[30 - x^{\frac{3}{2}} = t\]. Then \[-\frac{3}{2} \sqrt{x} dx = dt\] or \[\sqrt{x} dx = -\frac{2}{3} dt\]

Thus, \[\int \frac{\sqrt{x}}{(30 - x^{\frac{3}{2}})^2} dx = -\frac{2}{3} \int \frac{dt}{t^2} = \frac{2}{3} \left[ \frac{1}{t} \right] = \frac{2}{3} \left[ \frac{1}{(30 - x^{\frac{3}{2}})} \right] = F(x)\]

Therefore, by the second fundamental theorem of calculus, we have

\[\int_{0}^{\frac{\pi}{4}} \sin^3 2t \cos 2t dt\]

Solution:

Let \[I = \int_{0}^{\frac{\pi}{4}} \sin^3 2t \cos 2t dt\].

Consider \[\int \sin^3 2t \cos 2t dt\]

Put \[\sin 2t = u\] so that \[2 \cos 2t dt = du\] or \[\cos 2t dt = \frac{1}{2} du\]

So

Therefore, by the second fundamental theorem of integral calculus

• The theorem connects differentiation and integration.

• If \[A(x) = \int_{a}^{x} f(t) \, dt\], then \[A'(x) = f(x)\].

• If  F'(x) = f(x), then \[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\].

• The result is used to evaluate definite integrals quickly.

• The function should be continuous on the interval for direct use of the theorem.