Integration and the Fundamental Theorem of Calculus
Finding antiderivatives and evaluating definite integrals
Integration and the Fundamental Theorem of Calculus
Finding antiderivatives and evaluating definite integrals
Math - Grade 9-12
- 1
Find an antiderivative of f(x) = 6x^2.
Use the power rule for antiderivatives.
An antiderivative is F(x) = 2x^3 + C because the derivative of 2x^3 is 6x^2. - 2
Find the indefinite integral integral of (4x^3 - 2x + 5) dx.
The indefinite integral is x^4 - x^2 + 5x + C because each term is integrated separately. - 3
Evaluate integral from 0 to 3 of 2x dx.
Find an antiderivative first, then apply the limits.
The value of the integral is 9. An antiderivative of 2x is x^2, and x^2 evaluated from 0 to 3 gives 9 - 0 = 9. - 4
Evaluate integral from 1 to 4 of (x + 2) dx.
The value of the integral is 13. An antiderivative is (1/2)x^2 + 2x, and evaluating from 1 to 4 gives 16 - 3 = 13. - 5
Use the Fundamental Theorem of Calculus to find the derivative of F(x) = integral from 2 to x of (t^2 + 3) dt.
For F(x) = integral from a to x of f(t) dt, the derivative is f(x).
The derivative is F'(x) = x^2 + 3 because the Fundamental Theorem of Calculus says that the derivative of an accumulation function is the integrand evaluated at x. - 6
Use the Fundamental Theorem of Calculus to find the derivative of G(x) = integral from -1 to x of (5t^4) dt.
The derivative is G'(x) = 5x^4 because the derivative of the integral from a constant to x equals the integrand evaluated at x. - 7
Find the value of integral from 0 to 2 of (3x^2 - 4) dx.
Integrate each term, then substitute the bounds.
The value of the integral is 0. An antiderivative is x^3 - 4x, and evaluating from 0 to 2 gives (8 - 8) - 0 = 0. - 8
Find an antiderivative of f(x) = 1/x for x > 0.
An antiderivative is F(x) = ln(x) + C because the derivative of ln(x) is 1/x for x > 0. - 9
Evaluate integral from 1 to e of (1/x) dx.
Use the natural logarithm as the antiderivative.
The value of the integral is 1. An antiderivative of 1/x is ln(x), and evaluating from 1 to e gives ln(e) - ln(1) = 1 - 0 = 1. - 10
If F'(x) = 8x - 3, find one possible function F(x).
One possible function is F(x) = 4x^2 - 3x + C because the derivative of 4x^2 - 3x is 8x - 3. - 11
Evaluate integral from -2 to 2 of x^3 dx.
Notice the symmetry of the function and interval.
The value of the integral is 0. An antiderivative is (1/4)x^4, and evaluating from -2 to 2 gives 4 - 4 = 0. This also makes sense because x^3 is an odd function over a symmetric interval. - 12
Find the average value of f(x) = x^2 on the interval [0, 3].
The average value is 3. The average value formula gives (1/(3 - 0)) times the integral from 0 to 3 of x^2 dx = (1/3) times 9 = 3. - 13
A particle moves with velocity v(t) = 3t^2 meters per second for 0 less than or equal to t less than or equal to 2. Find the displacement from t = 0 to t = 2.
Displacement over an interval is the integral of velocity.
The displacement is 8 meters. Integrating velocity gives integral from 0 to 2 of 3t^2 dt = [t^3] from 0 to 2 = 8 - 0 = 8. - 14
Suppose H(x) = integral from 0 to x of (2t + 1) dt. Find H(3).
The value is H(3) = 12. Evaluating the integral gives [t^2 + t] from 0 to 3 = (9 + 3) - 0 = 12. - 15
Evaluate integral from 2 to 5 of 7 dx.
A constant function forms a rectangle under the curve.
The value of the integral is 21. The integral of a constant 7 over the interval from 2 to 5 is 7(5 - 2) = 21.