Complete Calculus Mathematics - from Beginner to Advanced

Explore antiderivatives by identifying primitives of 2x as x^2 + C, and relate derivatives to tangent slopes on the curves x^2 and x^2+1.

Apply the antiderivative rule: for x^n, the primitive is x^{n+1}/(n+1) plus a constant, for n ≠ -1. Use examples like x^2, x^3, and x^4 to illustrate exponent rise and division.

Explore the integral function as the primitive of a function, showing how integration undoes differentiation with F(x) plus a constant, and illustrate with 2x integrating to x^2 plus C.

Explore the general rule for integration, the power rule, which increases the exponent by one and divides by the new exponent, with examples like x^7/7.

Compute the constant of integration for a curve with slope 4x+5; using (2,3) yields C = -15, so y = 2x^2 + 5x - 15.

Solve dy/dx = 2x - 1 to get y = x^2 - x + C. Then use y(2)=5 to find C=3, yielding y = x^2 - x + 3.

Apply substitution to integrate a function of a function, raise the inner power by one, and divide by the inner derivative, using the ax+b to the power rule.

Integrate (3x-2)^3 dx to obtain (3x-2)^4/12 + C, and integrate dx over sqrt(5+6x) by substitution to obtain sqrt(5+6x)/3 + C.

Learn how to integrate trigonometric functions and reverse differentiation, using sin x, cos x, and composite forms like sin(2x+1) and cos(2x+1), plus sec^2(2x+1) giving tan(2x+1)/2.

Explore the general rules for integrating trig functions, with examples involving sine terms and how constants and denominators arise in the antiderivative.

Tackle trig integration with worked examples, deriving antiderivatives of sine and tangent forms and applying the constant of integration.

The definite integral measures the area under the curve above the x-axis between i and b, computed by evaluating the antiderivative at b and i.

Calculate the area under the curve x^2 from x=1 to x=3 by using primitives and the definite integral, applying the upper and lower limits to obtain the result.

Demonstrate definite integrals with worked examples: evaluate from 1 to 2 and 0 to 3 for polynomials and powers, apply antiderivatives, limit evaluation, and combine results for x^2 and 1/√x.

Explore how distance as a function of time differentiates to velocity, velocity differentiates to acceleration, and integrate acceleration to recover velocity and distance from time.

Learn how acceleration, the derivative of velocity with respect to time, leads to velocity by integration, and how v(0)=5 m/s fixes the integration constant.

Substitute t = 2 into the velocity expression to evaluate the rate of change, yielding three and two-thirds meters per second.

Apply calculus to a particle with given acceleration and initial velocity to determine distance traveled in the first two seconds, via integrating acceleration to get velocity and distance.

Apply calculus to relate velocity to distance by treating velocity as the derivative of distance with respect to time and integrating over time. With limits from 0 to 2 seconds, the distance traveled in the first two seconds is 12 meters.

Apply integration by substitution to compute the integral of (1+x^2) x dx using u = 1 + x^2, then back-substitute to obtain the final expression.

Apply the substitution u = sin x to evaluate the integral ∫ sin^2 x cos x dx, giving (1/3) sin^3 x + C.

Explore the application of integration by substitution in a square-root problem, derive the substitution, simplify, and back-substitute to obtain the integral.

Apply calculus to prove circle area by integrating y = sqrt(r^2 - x^2) from 0 to r, using x = r sin theta with limits 0 to pi/2.

Apply the r-squared integration to compute the area of a circle, identify the quarter-circle area, and show that multiplying by four yields the circle's area, involving pi.

Rotate the parabola y^2 = 4x about the x-axis from 0 to 1 using V = pi ∫ y^2 dx with y^2 = 4x, yielding 2 pi.

Explore volumes of solids of revolution by rotating curves about the y-axis and x-axis, using pi times the integral of squared functions, and convert between x- and y-expressions.

Explore differential calculus by finding gradients and tangents: differentiate y = x^2 − 3 to obtain gradient 2x, then substitute x = 1 to get the tangent's slope of 2.

Calculate the gradient at x=2 as 11, then use the point (2,6) and the point gradient form to derive the tangent: y = 11x - 16.

Compute the normal to y = x^3 + 7x^2 - 6 at x = 1, tangent slope 17 and normal slope -1/17, giving y - 2 = -(1/17)(x - 1).

Explore differential calculus by differentiating f(x)=20x^2 at x=2 to get 80, and apply PV=60 to find dP/dV=-60/V^2, -15 at V=2.

Explore how dy/dx and d^2y/dx^2 determine turning points in a curve. A negative second derivative signals a maximum, while a positive second derivative signals a minimum when dy/dx equals zero.

Find the minimum turning point of the curve y = 2x^2 - x by solving dy/dx = 0, verify with d^2y/dx^2 > 0, and substitute to get (1/4, -1/8).

Explore turning points with first and second derivatives, solving x^2-1=0 to x=±1, and classify a minimum at (1,-2) and a maximum at (-1,2).

Solve maxima and minima problems by dividing 50 into two parts to maximize their product; differentiate to find x=25, giving the maximum at 25 and 25.

Using the existing fence as one boundary, allocate 400 meters of new fencing for the other three sides with two equal sides x and boundary side y, giving area = x(400−2x) = 400x−2x^2. The derivative test shows a maximum at x = 100 m, y = 200 m, yielding a maximum area of 20000 square meters.

Form an open box from a 20 cm square by cutting out squares of side x, giving V(x)=x(20−2x)^2; differentiate to locate the maximum or minimum.

Apply differential calculus to locate maxima and minima using the second derivative; identify a maximum at x ≈ 3.33 cm and compute the volume ≈ 592.6 cm³.

Identify inflection points in differential calculus by locating where the second derivative is zero and changes sign, as shown on the curve y = -6x^2 + 5x.

Compute the second derivative, set it to zero to locate the inflection point at x = 2, with y = -2 on the curve, where the second derivative changes sign.

Explore the second derivative in differential calculus by differentiating the first derivative, with examples like y = -x^2 and y = x^4 + x^3.

Compute the area under a curve above the x-axis from 0 to 2 by evaluating the definite integral of x^4/4.

Negative areas occur when region lies below x axis; area under y equals x^4 from -2 to 0 is negative, yet the actual area is 4 square units.

Compute the area between the curve x^2 - 1 and the x-axis from 0 to 1 by taking the absolute value of the definite integral, yielding 2/3 units squared.

Compute the area between the curve x(x-1)(x-3) and the x-axis by integrating 0 to 1 (positive region) and using the absolute value on 1 to 3 (negative region).

Explore area under curves using integrals and the top limit to compute area, as taught in lesson 5 of area and volume, yielding square units.

Compute volumes of revolution about the x axis by rotating y=f(x). Use V = pi ∫ [f(x)]^2 dx from 0 to 2; e.g., y=x^2 gives 32/5 pi.

Compute volumes of revolution about the y axis by rotating a curve and expressing it as x equals a function of y, then apply V = pi ∫ f(y)^2 dy.

Continue rotating the curve about the y axis to find the volume of revolution. The calculation yields a volume of 8 pi units.

Derive a sphere’s volume via volumes of revolution by integrating y^2 = r^2 − x^2, obtaining 2/3 pi r^3 for the hemisphere and 4/3 pi r^3 for the sphere.