\(\int \sqrt x(3x^2+2x+3)dx\)
= \(\int \bigg(3x^{\frac{5}{2}}+2x^{\frac{3}{2}}+3x^{\frac{1}{2}}\bigg)dx\)
=\(3 \int x^{\frac{5}{2}}dx+2 \int x^{\frac{3}{2}}dx+3 \int x^{\frac{1}{2}}dx\)
=\(3\bigg(\frac{x^{\frac{7}{2}}}{\frac{7}{2}}\bigg)+2\bigg(\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\bigg)+3\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+C\)
=\(\frac{6}{7}x^{\frac{7}{2}}+\frac{4}{5}x^{\frac{5}{2}}+2x^{\frac{3}{2}}+C\)
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: