\(\int \frac{x^3+5x^2-4}{x^2}dx\)
=\(\int (x+5-4x^{-2})dx\)
=\(\int xdx +5 \int 1.dx-4\int x^{-2}dx\)
=\(\frac{x^2}{2}+5x-4\bigg(\frac{x^{-1}}{-1}\bigg)+C\)
=\(\frac{x^2}{2}+5x+\frac{4}{x}+C\)
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Integral calculus helps to resolve two major types of problems: