Set up and evaluate the definite integral that gives the area of the region bounded by the graph

Set up and evaluate the definite integral that gives the...

Key Concepts

Definite Integrals

A definite integral is used to compute the accumulation of quantities, such as area under a curve, between two specified limits. It is a fundamental tool in calculus that evaluates the integral from a lower bound to an upper bound, providing the sum of infinitely small quantities over the interval.

Area Between Curves

Calculating the area between curves involves setting up an integral where the integrand is the difference between the functions describing the upper and lower boundaries of the region. Determining which function is on top over the interval of interest is essential, and the integral is taken over the range bounded by the points of intersection of the curves.

Tangent Line

The tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. Its slope is equal to the derivative of the function at that point. Tangent lines are used to approximate the behavior of functions locally and are critical in setting up problems involving areas between curves and their tangents.

Derivatives

The derivative of a function measures the rate at which the function's value changes with respect to changes in the input variable. It is used to determine the slope of the tangent line at a point on the curve, which in turn is essential for constructing the equation of the tangent line.

Intersection Points

Intersection points between the graph of a function and its tangent line (or between any two curves) act as the limits of integration when calculating the area of the region between them. Finding these points typically involves equating the functions and solving for the variable, which ensures that the entire region of interest is captured in the integral.

Transcript

00:01 In this question, we have a function, y is equal to x cube minus 2x and a tangent line to this function passing through point minus 1 .1.

00:14 We are required to find the area of the region bounded by the graphs of y and the tangent line passing through point minus 1 .1.

00:24 So let's see how to solve this question.

00:27 First of all, let's find the equation of the tangent line.

00:34 The equation y is equals to x cube minus 2x and now let's differentiate this so we can write y dash is equals to 3x square minus 2 now let's find the slope of the tangent line at minus 1 comma 1 so substitute x is equal to minus 1 in the above equation so we we get y -dash is equal to 3 into minus 1 to the power 2 minus 2.

01:13 So from here we will have y -dash is equals to 1.

01:19 So this is the slope of the line.

01:22 We know that the standard equation of line can be written as y -minus y1 is equal to slope y -dash x minus x1.

01:33 Now, substitute all the values.

01:37 So we get y -minus 1 is equal to the 1.

01:41 To 1 into x minus minus 1.

01:48 So from here we will have y is equals to x plus 2.

01:55 So this is the equation of tangent line.

01:59 And now let's find the another point of intersection and to do so equate x cube minus 2x and x plus 2.

02:11 So from here we get the equation x cube minus 3x minus 2 is equal to 0.

02:25 When we solve this we get x is equal to 2.

02:33 So now on the basis of these data, let's draw the graph for y and the tangent line...