### Home > CALC > Chapter 8 > Lesson 8.2.3 > Problem8-87

8-87.

$\text{Since }g(x)\text{ was defined as }\int_{0}^{x}f(t)dt,$

$g(2)=\int_{0}^{2}f(t)dt$

$g(4)=\int_{0}^{4}f(t)dt$

Consider the symmetry of the f(t) between t = −2 and t = 2.

$\int_{-2}^{2}f(t)dt=kg(a)$

What is k? What is a?

Points of NON-differentiablity include cusps, endpoints, jumps, holes and vertical tangents.

Recall that a local maximum exists where the derivative changes from positive to negative. This could happen where f '(x) = 0 or where f '(x) = DNE.

Notice that g'(x) = f(x); after all, the derivative of an integral is the original function.

What is the slope of g(x) at x = 4 (see second hint in part (e))? What is the y-value?

Concavity is the slope of the slope. So inflection points are where the slope of the slope changes.