Question

If the area function ${\int }_a^{x}f\left(x\right)dx=x^2-a^2$ then f(x) =

• A. $\frac{x^3}{3}$
• B. $2x$
• C. $3x^2$

Answer 1

The correct option is B $2x$

We have defined ${\int }_a^{b}f\left(x\right)dx$ as the area of region bounded by curve f(x), ordinates x = a & x = b and the x- axis.

Let x be a given point in [a,b] then ${\int }_a^{x}f\left(x\right)dx$ represents the area of the light shaded portion. The area of this shaded portion depends on x. In other words , the area of this shaded portion is the function of x. We call this function area function and represents it with A(x).
So, $A\left(x\right)={\int }_a^{x}f\left(x\right)dx$
Now from first fundamental theorem of integral calculus we can say
A’(x) = f(x) .
From the question given if we compare we get $A\left(x\right)=x^2-a^2$
So, A’(x) = 2x
Also A’(x) = f(x) (from first fundamental theorem of integral calculus)
$\Rightarrow f\left(x\right)=2x$