Calculus: Integrals
Finding antiderivatives, definite integrals, and area under curves
Calculus: Integrals
Finding antiderivatives, definite integrals, and area under curves
Calculus - Grade 9-12
- 1
Evaluate the indefinite integral: ∫ 6x^2 dx.
Use the power rule: ∫ x^n dx = x^(n+1)/(n+1) + C for n not equal to -1.
The integral is 2x^3 + C because the power rule gives ∫ 6x^2 dx = 6(x^3/3) + C = 2x^3 + C. - 2
Evaluate the indefinite integral: ∫ (4x^3 - 5x + 7) dx.
The integral is x^4 - (5/2)x^2 + 7x + C. Each term can be integrated separately using the power rule. - 3
Evaluate the definite integral: ∫ from 0 to 3 of 2x dx.
Find an antiderivative first, then subtract the value at the lower limit from the value at the upper limit.
The value is 9. An antiderivative of 2x is x^2, so x^2 evaluated from 0 to 3 is 3^2 - 0^2 = 9. - 4
Evaluate the definite integral: ∫ from 1 to 4 of x^2 dx.
The value is 21. An antiderivative of x^2 is x^3/3, so the integral equals 4^3/3 - 1^3/3 = 64/3 - 1/3 = 63/3 = 21. - 5
Find the antiderivative of f(x) = 1/x for x > 0.
The power rule does not work when the exponent is -1.
An antiderivative is ln(x) + C because the derivative of ln(x) is 1/x for x > 0. - 6
Evaluate the indefinite integral: ∫ (3e^x + 2) dx.
The integral is 3e^x + 2x + C because the antiderivative of e^x is e^x and the antiderivative of 2 is 2x. - 7
Evaluate the definite integral: ∫ from 0 to π of sin(x) dx.
Remember that cos(0) = 1 and cos(π) = -1.
The value is 2. An antiderivative of sin(x) is -cos(x), so the integral equals -cos(π) - [-cos(0)] = 1 - (-1) = 2. - 8
Evaluate the indefinite integral: ∫ cos(x) dx.
The integral is sin(x) + C because the derivative of sin(x) is cos(x). - 9
A velocity function is v(t) = 5t meters per second for 0 ≤ t ≤ 4. Find the displacement from t = 0 to t = 4.
Displacement is the integral of velocity over time.
The displacement is 40 meters. The displacement is ∫ from 0 to 4 of 5t dt = (5/2)t^2 evaluated from 0 to 4, which is (5/2)(16) = 40. - 10
Use geometry to evaluate ∫ from 0 to 6 of 3 dx.
A constant function creates a rectangle under the curve.
The value is 18. The graph y = 3 forms a rectangle with width 6 and height 3, so the area is 6 × 3 = 18. - 11
Evaluate the indefinite integral: ∫ (8x^7 - 6x^2 + 4) dx.
The integral is x^8 - 2x^3 + 4x + C because ∫ 8x^7 dx = x^8, ∫ -6x^2 dx = -2x^3, and ∫ 4 dx = 4x. - 12
Evaluate the definite integral: ∫ from -1 to 1 of x^3 dx.
An odd function has symmetry about the origin. Equal intervals on both sides of 0 can cancel in a definite integral.
The value is 0. The function x^3 is odd, so the signed areas from -1 to 0 and from 0 to 1 cancel each other. - 13
Find the average value of f(x) = x^2 on the interval [0, 3].
Average value on [a, b] is (1/(b - a))∫ from a to b of f(x) dx.
The average value is 3. The formula is (1/(3 - 0))∫ from 0 to 3 of x^2 dx = (1/3)(x^3/3 evaluated from 0 to 3) = (1/3)(9) = 3. - 14
Evaluate the indefinite integral: ∫ 1/(x^2) dx for x not equal to 0.
The integral is -1/x + C. Rewrite 1/(x^2) as x^-2, then use the power rule to get x^-1/(-1) + C, which is -x^-1 + C. - 15
The graph of f(x) is above the x-axis on [0, 2] and forms a triangle with base 2 and height 5. What is ∫ from 0 to 2 of f(x) dx?
When the graph is above the x-axis, the definite integral equals the area under the curve.
The value of the integral is 5. Since the region is a triangle above the x-axis, the integral equals the area, which is (1/2)(2)(5) = 5.