Find exact areas and assess numerical integration results

Reuse area and numerical foundations

Here logarithms are natural and angles are radians. A complete solution identifies the region and domain before choosing exact or numerical integration

Signed integral and geometric area answer different questions

A definite integral adds signed vertical contributions. Geometric area adds their magnitudes. For a continuous f, area between f and the axis is ∫|f(x)| dx; between f and g it is ∫|f(x)−g(x)| dx. Find sign changes or intersections, compare heights on each interval and add positive pieces

For the shown odd integrand on [−1,1], which statement is correct?

Use an inverse-sine primitive on a proper interval

For f(x)=3/√(16−(x−1)²) on [−1,3], use F(x)=3arcsin((x−1)/4). The endpoint inverse inputs are −1/2 and 1/2, so the integral is 3(π/6+π/6)=π. The integrand is positive and the interval lies inside −3<x<5, so this is also the geometric area

Give the coefficient of π in the geometric area under this curve

A negative curve has a negative integral and positive area

For f(x)=−6/[9+(x+1)²] on [−4,2], a primitive is −2arctan((x+1)/3). The signed integral is −2(π/4+π/4)=−π. Since f is negative throughout, geometric area is π. The absolute value shortcut is valid here because the sign does not change

Give the coefficient of π in the signed integral of this function, not the geometric area

Find intersections before subtracting a line

For y=2/(1+x²) and y=1, solve 2/(1+x²)=1 to get x=−1,1. Between these values the curve lies above the line. The bounded area is ∫ from −1 to 1 of [2/(1+x²)−1] dx

Which width describes the bounded region in this graph?

Evaluate the cap exactly

The primitive 2arctan x−x gives (π/2−1)−(−π/2+1)=π−2, approximately 1.14159. Retain π−2 as the exact answer. Its decimal is a separate approximation; its positivity agrees with the region

For the bounded region between these curves, write its area as Pπ+R and enter R

Horizontal strips use right minus left

When boundaries are naturally written x=R(y) and x=L(y), a horizontal width is R(y)−L(y). Integrate with respect to y between the intersection y-values. A curve can fail to be one function y=f(x) while still being a useful x-function of y.

Work a horizontal-strip region

For x=y²/2 and x=3y, the intersections satisfy y²/2=3y, giving y=0,6. On [0,6], the line is the right boundary. Area is ∫ from 0 to 6 of (3y−y²/2)dy=[3y²/2−y³/6] from 0 to 6=18.

What is the upper y-limit for horizontal-strip integration of the region bounded by these curves?

Use exact primitives when they are available

Keep a symbolic expression until a decimal is requested. For the region under y=x e^(−x) on [0,2], parts gives primitive −(x+1)e^(−x), hence exact area 1−3e^(−2). The curve is nonnegative and the result is approximately 0.593994. Check the derivative, limits and sign separately

Numerical integration does not require a displayed primitive

A continuous function such as e^(−x²) on [0,1] has a well-defined definite integral. Numerical quadrature can approximate it even if a symbolic tool leaves it unevaluated. Input the full expression and bounds, use radians where needed, retain working precision and record the method or settings

Trapezia with equal spacing

For equal width h=(b−a)/n, add adjacent average heights. The endpoint heights occur once and interior heights twice before the factor h/2. This is also the mean of the left and right rectangle sums on the same partition

Unequal spacing needs individual widths

For sampled values (x,f(x))=(0,1),(1,3),(3,2),(4,4), the gaps are 1,2,1. The three trapezia contribute 1(1+3)/2=2, 2(3+2)/2=5 and 1(2+4)/2=3, giving T=10. One common width would misweight the middle interval

Find the trapezium contribution from x=1 to x=3 using these endpoint heights

State what the table means

A trapezium sum is exactly the signed integral of the piecewise-linear interpolant through the samples. It is generally only an approximation to an unknown curved function with those same samples. A finite table alone gives no universal error bound; narrow features can be missed between sample inputs

An interpolating segment can cross the axis

Define f by straight segments through (0,3),(2,−1),(5,2). The first zero is 3/2 and the second is 3. The signed trapezium sum is 2(3−1)/2+3(−1+2)/2=7/2. Four positive triangles give geometric area 9/4+1/4+1/2+2=5.

A continuous piecewise-linear function joins these points in order; what is its geometric area over the displayed interval?

Monotonicity and concavity give different information

For an increasing continuous function, left endpoint sums are no greater than the signed integral and right sums are no less. A trapezium chord lies above a convex graph and below a concave graph. Increasing alone does not decide the trapezium error direction. For a line the trapezium rule is exact

Which information justifies calling a trapezium sum an upper bound?

Agreement is a check, not an accuracy certificate

A finer partition or tighter requested tolerance often improves a calculation, but two agreeing outputs can share a bias or miss the same feature. A tool tolerance is a requested control, not automatically a proved absolute error bound. Use independent symbolic checks, domain analysis or a justified error enclosure when claiming certified digits

A supplied enclosure can certify a rounded answer

Suppose a separate rigorous argument proves 1.2501≤I≤1.2504. Both endpoints round to 1.250 at three decimal places, so every enclosed value does too. If instead 1.25049≤I≤1.25051, the endpoints round differently and that interval alone cannot fix the three-decimal result

Enter 1 if this rigorous enclosure certifies I rounded to three decimal places, otherwise 0

A domain failure is not a rounding problem

For 1/(x−2) on [1,3], the two one-sided improper integrals diverge at 2. Endpoint logarithms cancelling to 0 or a numerical routine returning 0 does not make the ordinary integral finite. A symmetric principal value is a different calculation. Check the original interval, not just a graph window

Set up and compare strip directions, ungraded

Find the region bounded by x=y² and x=y+2. Sketch its boundaries, calculate its area with horizontal strips, and set up an equivalent vertical-strip calculation. Explain why the vertical calculation needs two pieces

Model: one horizontal interval, two vertical intervals

The intersection y-values are −1 and 2. Right minus left gives ∫ from −1 to 2 of (y+2−y²)dy=9/2. With vertical strips, the region extends from x=0 to 4. For 0≤x≤1, its lower branch is −√x and upper branch √x. For 1≤x≤4, the lower boundary is x−2 and the upper is √x. Thus ∫ from 0 to 1 of 2√x dx+∫ from 1 to 4 of (√x−x+2)dx=9/2. A complete labelled learner sketch remains ungraded

Justify exact and numerical results, ungraded

For the cap between 2/(1+x²) and 1, derive the exact area π−2, then use symbolic differentiation and numerical integration to verify it. Record your actual tool inputs, angle mode, domain checks, precision settings and output. Explain why the tool alone does not identify the intended region

Model: check the setup before the computation

The roots −1 and 1 and height comparison establish the region. Differentiating 2arctan x−x verifies the integrand. A numerical definite evaluation on [−1,1] gives about 1.1415926536. Using the curve alone would produce π, a different region. This supplied output is a model, not evidence that the learner used technology

Investigate a numerical integral, ungraded

Use a numerical integration tool for ∫ from 0 to 1 of e^(−x²)dx. Compare results with a refined partition or tighter tolerance and with a second method if available. Report a sensible decimal approximation, what checks you performed, and which accuracy claims you cannot prove from agreement alone

Model: separate existence, approximation and error

The integrand is continuous and positive on [0,1]. A numerical result is approximately 0.7468241328. Endpoint monotonicity gives useful coarse bounds because e^(−x²) decreases on this interval. Refinement and method agreement are evidence of stability, not a rigorous ten-decimal certificate. Do not write the displayed decimal as an exact equality or claim a tool was executed merely by reading this model

Prove the table limitation and trapezium identity, ungraded

Derive the trapezium contribution for one straight segment. Explain why a finite table cannot determine the exact integral of every continuous function through its points. Give two such functions with different integrals

Model: interpolation does not fix the curve

A straight segment has signed integral h(y0+y1)/2 by integrating its linear equation. For nodes 0 and 1 with both heights 0, f(x)=0 and g(x)=x(1−x) share the samples but have integrals 0 and 1/6. Multiplying the second function by any positive constant changes the integral while keeping the same samples. Extra regularity information, such as a derivative bound, is needed for a certified approximation error

Explain signs, singularities and certified digits, ungraded

Explain why ∫|f| generally differs from |∫f|, why a pole must be checked even if endpoint values look harmless, and why a rigorous enclosure can justify rounding while two agreeing approximations cannot. Use one concrete example of each

Model: give separate counterexamples and hypotheses

For the piecewise-linear example, area 5 differs from |7/2|. For 1/(x−2) on [1,3], endpoint logarithm cancellation hides divergent one-sided integrals. For a rigorous enclosure [1.2501,1.2504], every value rounds to 1.250 at three decimals. Agreement of two estimates supplies no such all-values statement without an error argument