Study Guide12th GradeLevel 3 · Challenge

Free 12th Grade Integration and Area Under Curves

A senior study guide to definite integrals and area: the standard integrals, the fundamental theorem of calculus, areas below the x-axis, areas between two curves and the trapezoidal rule. Each worked example shows the antiderivative in square parentheses, substitutes the limits and interprets the sign, with a six-question self-check and answers.

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Grade: 12th Grade
Subject: Math
Topic: Calculus
Difficulty: Level 3 · Challenge
Estimated time: 30 minutes
Curriculum: Texas grade levels (TEKS)
Answers: Not applicable
Format: PDF (A4) + print

Students will practice

finding antiderivatives of polynomial, linear-power, exponential, reciprocal and trigonometric functions
evaluating definite integrals using the fundamental theorem of calculus
finding the area between a curve and the x-axis, including regions below the axis
finding the area between two curves and approximating area with the trapezoidal rule

Curriculum: Texas grade levels (TEKS). We show specific outcome codes only where they have been verified against the official curriculum document.

What's next?

Completed: Integration and Area Under Curves

Study guide

How to use this study guide

Read it together first, pausing at each worked example to try the step before reading the answer.
Attempt the "Check yourself" questions at the end without looking back.
Then practice with a worksheet from the pathway above and finish with the topic test.

Common questions

Who is this study guide for?

12th Grade students (typically ages 17–18) working on calculus. It is pitched at level 3 · challenge.

Are the answers included?

This is a study guide, so there is no separate answer sheet; the 'Check yourself' questions include answers.

How long does it take?

About 30 minutes. Short, regular sessions work best: two or three a week beats one long one.

Do I need to sign up to download?

No. Click Download Free PDF and it opens immediately. It is free for personal, classroom and homeschool use.

What should we do next?

Try Differentiating Exponential, Logarithmic and Trigonometric Functions.

Success TutoringWorksheets12th Grade · Mathematics · Calculus

Integration and Area Under Curves

What you need to know

Integration reverses differentiation. An indefinite integral ∫ f(x) dx is a family of functions F(x) + C whose derivative is f(x). A definite integral has limits: by the fundamental theorem of calculus, ∫ from a to b of f(x) dx = F(b) − F(a), written [F(x)] with the limits b and a. When f(x) ≥ 0 on the interval, this number is the area between the curve and the x-axis from x = a to x = b.

Function f(x)	Antiderivative F(x)	Note
xⁿ (n ≠ −1)	+ C	raise the power by 1, divide by the new power
(ax + b)ⁿ (n ≠ −1)	+ C	also divide by the coefficient of x
e^ax	e^ax + C	e^x is its own integral
	ln \|x\| + C	the n = −1 case
cos x	sin x + C	x in radians
sin x	−cos x + C	note the sign

Signed area. Where the curve is below the x-axis the integral is negative. To find a physical area, integrate each part separately and add the absolute values. Area between two curves: find where they intersect, then integrate (top curve − bottom curve) between those x-values — this works even if part of the region is below the axis.

Evaluate ∫ from 0 to 2 of (3x² + 2x) dx

Antiderivative: [x³ + x²] from 0 to 2
= (2³ + 2²) − (0 + 0) = 8 + 4 = 12

Find ∫ (2x + 1)⁴ dx and ∫ from 0 to 1 of e^2x dx

∫ (2x + 1)⁴ dx = + C = [frac (2x + 1)[sup 5]/10] + C
∫ from 0 to 1 of e^2x dx = [e^2x] from 0 to 1 = e² − e⁰ = [frac e[sup 2] − 1/2] ≈ 3.19

Find the area between y = x² − 4 and the x-axis from x = 0 to x = 2

On 0 ≤ x ≤ 2 the curve is below the axis (for example y = −4 at x = 0), so expect a negative integral.
∫ from 0 to 2 of (x² − 4) dx = [ − 4x] from 0 to 2 = ( − 8) − 0 = −
The integral is −, so the area is [frac 16/3] square units (about 5.33). State the area as a positive number.

Find the area enclosed between y = x + 2 and y = x²

Intersections: x² = x + 2, so x² − x − 2 = 0, (x − 2)(x + 1) = 0, giving x = −1 and x = 2.
Between these, the line is on top (test x = 0: line gives 2, parabola gives 0).
Area = ∫ from −1 to 2 of (x + 2 − x²) dx = [ + 2x − ] from −1 to 2
At x = 2: 2 + 4 − = . At x = −1: − 2 + = −.
Area = − (−) = + = = [frac 9/2] square units

Trapezoidal rule: approximate ∫ from 0 to 2 of √(1 + x³) dx using 2 subintervals

h = = 1. Function values: f(0) = √1 = 1, f(1) = √2 ≈ 1.4142, f(2) = √9 = 3.
Trapezoidal rule: ∫ ≈ [f(0) + 2f(1) + f(2)] = [1 + 2(1.4142) + 3]
= × 6.8284 = 3.41 (to 2 decimal places)
The function is concave up here, so the trapezoids slightly overestimate the true area.

Common mistake: forgetting to divide by the coefficient of x when integrating (ax + b)ⁿ or e^ax. Check by differentiating your answer: of = = (2x + 1)⁴. Correct.

Common mistake: integrating straight across an x-intercept. If the curve crosses the axis inside the interval, the positive and negative parts partly cancel and the integral is smaller than the area. Find the intercepts first and split the integral there.

Check yourself

Find ∫ (4x³ − 2x) dx.
Evaluate ∫ from 0 to 3 of 2x dx.
Find ∫ (3x − 2)² dx.
Evaluate ∫ from 0 to 1 of e^x dx, exactly and to 2 decimal places.
Find the area between y = x³ and the x-axis from x = −1 to x = 0.
Find the area enclosed between y = 4 − x² and the x-axis.

Answers: 1. x⁴ − x² + C 2. [x²] from 0 to 3 = 9 3. + C 4. e − 1 ≈ 1.72 5. ∫ = [] from −1 to 0 = 0 − = −, so the area is square units 6. x-intercepts ±2; ∫ from −2 to 2 of (4 − x²) dx = [4x − ] from −2 to 2 = (8 − ) − (−8 + ) = square units

Preview — the PDF contains the full resource.

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Created by: Success Tutoring
Last reviewed: 1 October 2026
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