Integration and Computation — Comprehensive Exercises

date2026-07-15document_iddoc_0b51cb35fb722b4012853008388ac3a1descriptionリーマン和、符号付き面積、微分積分学の基本定理、置換積分、部分積分を横断して確認する総合演習である。prerequisites積分の定義：リーマン和と符号付き面積 / 原始関数と不定積分 / 微分積分学の基本定理 / 積分公式と計算法type問題演習content_typeexercisestatusactive

mathcalculusexerciseintegralfundamental-theorem

data/lecture/math/calculus/integral-definition-riemann-sums-and-signed-area.lecture.n.md data/lecture/math/calculus/fundamental-theorem-of-calculus.lecture.n.md data/lecture/math/calculus/integration-rules-and-computation.lecture.n.md

1Exercise Objectives

These exercises review the definition of a definite integral, signed area, differentiation with variable endpoints, substitution, and integration by parts. For each problem, identify the definition or theorem being used and state the conditions required for its application, rather than reporting only the numerical result.

2Problem 1

Partition $[0, 1]$ into $n$ equal subintervals and use the right endpoint of each subinterval as the sample point. Evaluate

\int_{0}^{1} x^{2} d x

from the resulting Riemann sum. You may use

\sum_{i = 1}^{n} i^{2} = \frac{n (n + 1) (2 n + 1)}{6.}

2.1Sample Solution

Let $f (x) = x^{2}$ . The function $f$ is continuous, hence Riemann integrable, on $[0, 1]$ . The subinterval width is $Δ x = 1 / n$ , and the right endpoint is $x_{i} = i / n$ . Therefore, the Riemann sum is

\sum_{i = 1}^{n} f (x_{i}) Δ x = \frac{1}{n^{3}} \sum_{i = 1}^{n} i^{2} = \frac{n (n + 1) (2 n + 1)}{6 n^{3}} .

Consequently,

\int_{0}^{1} x^{2} d x = \lim_{n \to \infty} \frac{n (n + 1) (2 n + 1)}{6 n^{3}} = \frac{1}{3.}

2.2Explanation

Each term of a Riemann sum is the product of a function value $f (x_{i})$ and a subinterval width $Δ x$ . Omitting $Δ x$ would sum function values rather than area elements.

3Problem 2

Evaluate

\int_{- 1}^{2} (x - 1) d x

and the geometric area bounded by $y = x - 1$ , the $x$ -axis, $x = - 1$ , and $x = 2$ .

3.1Sample Solution

The function $x - 1$ is continuous on $[- 1, 2]$ , so the fundamental theorem of calculus applies. The function $F (x) = (x - 1)^{2} / 2$ is an antiderivative of $x - 1$ . Hence,

\int_{- 1}^{2} (x - 1) d x = {[\frac{(x - 1)^{2}}{2}]}_{{- 1}^{2}} = \frac{1}{2} - 2 = - \frac{3}{2.}

In contrast, $x - 1$ changes sign at $x = 1$ , so the geometric area is

- \int_{- 1}^{1} (x - 1) d x + \int_{1}^{2} (x - 1) d x = 2 + \frac{1}{2} = \frac{5}{2.}

3.2Explanation

A definite integral represents signed area and counts contributions below the $x$ -axis as negative. To obtain geometric area, split the interval at each zero and add the absolute value of each contribution.

4Problem 3

Let

H (x) = \int_{x}^{x^{2}} \sin t d t .

Find $H^{'} (x)$ .

4.1Sample Solution

The integrand $\sin t$ is continuous, and the endpoint functions $x$ and $x^{2}$ are differentiable, so the variable-endpoint formula applies. Introducing the fixed endpoint $0$ gives

H (x) = \int_{0}^{x^{2}} \sin t d t - \int_{0}^{x} \sin t d t .

The fundamental theorem of calculus and the chain rule yield

H^{'} (x) = 2 x \sin (x^{2}) - \sin x .

4.2Explanation

The upper endpoint $x^{2}$ contributes $\sin (x^{2}) \cdot 2 x$ , while the lower endpoint $x$ contributes $- \sin x$ . The lower-endpoint term has a minus sign, and each endpoint contribution is multiplied by the derivative of that endpoint.

5Problem 4

Select an appropriate integration method and evaluate

\int_{0}^{1} 3 x^{2} e^{x^{3}} d x .

State the reason for your choice and the transformed interval of integration.

5.1Sample Solution

The function $g (x) = x^{3}$ is continuously differentiable on $[0, 1]$ , and $e^{u}$ is continuous on $g ([0, 1]) = [0, 1]$ . Moreover, the integrand has the form $e^{g (x)} g^{'} (x)$ , so substitution is appropriate. Let $u = x^{3}$ , so $d u = 3 x^{2} d x$ ; the transformed endpoints remain $0$ and $1$ . Therefore,

\int_{0}^{1} 3 x^{2} e^{x^{3}} d x = \int_{0}^{1} e^{u} d u = [e^{u}]_{0^{1}} = e - 1.

5.2Explanation

The integrand contains the composite function $e^{x^{3}}$ and the derivative $3 x^{2}$ of the inner function $x^{3}$ as factors. After changing the variable of a definite integral to $u$ , transform the endpoints to values of $u$ as well.

6Problem 5

Select an appropriate integration method and evaluate

\int_{0}^{1} x \sin x d x .

State the reason for your choice.

6.1Sample Solution

Take $u = x$ and $v^{'} = \sin x$ , so that $u^{'} = 1$ and $v = - \cos x$ ; their derivatives are continuous on $[0, 1]$ . Differentiating the polynomial simplifies it, so integration by parts is appropriate. Therefore,

\int_{0}^{1} x \sin x d x = [- x \cos x]_{0^{1}} + \int_{0}^{1} \cos x d x = - \cos 1 + \sin 1.

6.2Explanation

The polynomial factor $x$ simplifies to $1$ after differentiation, while $\sin x$ has the readily available antiderivative $- \cos x$ . For a definite integral, evaluate both the boundary term $[u v]_{0^{1}}$ and the remaining integral.

7Basic Exercises

data/exercise/math/calculus/substitution-and-integration-by-parts.exercise.n.md

8Related Lectures

data/lecture/math/calculus/antiderivatives-and-indefinite-integrals.lecture.n.md data/lecture/math/calculus/fundamental-theorem-of-calculus.lecture.n.md data/lecture/math/calculus/integration-rules-and-computation.lecture.n.md