Linearity and signed contributions for a sum of terms, integrals from easy to advanced lesson

The function $3x^2 - 4x + 2$ is a sum of three terms, and one of them, $-4x$ , is negative at every point of the interval $[0, 2]$ except $x = 0$ . Can its integral from $0$ to $2$ be found one term at a time, and what does a negative term do to the total? This lesson states the sum rule and the constant multiple rule that allow the split, keeps the sign of every piece, and finds the value $4$ . One antiderivative confirms it, and the average height of the graph gives a sense of its size.

Key terms

Sum rule for integrals

Definition Let $f$ and $g$ be integrable on $[a, b]$ , for example continuous there. Then $f + g$ and $f - g$ are integrable on $[a, b]$ , and

\begin{aligned} &\int_a^b \bigl(f(x) + g(x)\bigr)\,dx \\[0.4em] &\quad = \int_a^b f(x)\,dx \\[0.4em] &\qquad + \int_a^b g(x)\,dx, \end{aligned}

with a minus sign in both places for the difference $f - g$ . The reason lies in the Riemann sums. For every partition of $[a, b]$ and every choice of sample points $x_i^{*}$ , each strip of $f + g$ splits as

\begin{aligned} &\bigl(f(x_i^{*}) + g(x_i^{*})\bigr)\,\Delta x_i \\[0.4em] &\quad = f(x_i^{*})\,\Delta x_i \\[0.4em] &\qquad + g(x_i^{*})\,\Delta x_i, \end{aligned}

so every Riemann sum of $f + g$ is a Riemann sum of $f$ plus the matching Riemann sum of $g$ , and the limits add. The rule also holds with the limits of integration in the other order, because reversing them changes the sign of all three integrals. The rule needs both pieces to be integrable, and a sum can have an integral when its pieces do not. For example, the function $1/x - 1/(x + 1)$ has a finite improper integral from $1$ to $\infty$ , equal to $\ln 2$ , while neither $1/x$ nor $1/(x + 1)$ does.

In plain words When each piece has an integral of its own, integrate the pieces over the same interval and add the results. The addition happens height by height, the height of $f + g$ at a point $x$ being the height of $f$ plus the height of $g$ at that same $x$ . Placing two regions side by side over adjacent intervals is a different rule, the one that adds integrals over $[a, b]$ and $[b, c]$ .

Example The region under $y = x + 1$ on $[0, 2]$ is the rectangle of height $1$ with the triangle under $y = x$ stacked on top of it. The sum rule adds their integrals,

\begin{aligned} &\int_0^2 (x + 1)\,dx \\[0.4em] &\quad = \int_0^2 x\,dx + \int_0^2 1\,dx \\[0.4em] &\quad = 2 + 2 = 4. \end{aligned}

Geometry agrees. The region is a trapezoid of width $2$ with parallel sides $1$ and $3$ , so its area is $\tfrac12(1 + 3) \cdot 2 = 4$ .

Constant multiple rule

Definition Let $f$ be integrable on $[a, b]$ and let $c$ be a constant, a real number that does not depend on the variable of integration. Then $cf$ is integrable on $[a, b]$ , and

\begin{aligned} &\int_a^b c\,f(x)\,dx \\[0.4em] &\quad = c\int_a^b f(x)\,dx. \end{aligned}

Each term $c\,f(x_i^{*})\,\Delta x_i$ of a Riemann sum of $cf$ carries the factor $c$ , so every Riemann sum of $cf$ is $c$ times the matching Riemann sum of $f$ , and the limits differ by the same factor.

In plain words A constant factor can move outside the integral, because it multiplies the height of every strip by the same number and leaves the widths alone. A negative constant also reflects every strip across the axis, so it changes the sign of the integral. A factor that changes with $x$ cannot come out. For example $\int_0^1 x \cdot x\,dx = \tfrac13$ , while pulling one $x$ outside would give the expression $x \int_0^1 x\,dx = x/2$ , which depends on $x$ and is not a number at all.

Example The power rule gives $\int_0^2 x^2\,dx = \tfrac83$ , so

\int_0^2 3x^2\,dx = 3 \cdot \tfrac83 = 8.

Every strip under the graph of $3x^2$ is three times as tall as the matching strip under $x^2$ and just as wide, so the area is three times as large. One antiderivative of $3x^2$ is $x^3$ , and $2^3 - 0^3 = 8$ agrees. With the constant $-1$ the rule gives $\int_0^2 (-x^2)\,dx = -\tfrac83$ , the signed value of a region of area $\tfrac83$ below the axis.

Linearity of the integral

Definition Let $f$ and $g$ be integrable on $[a, b]$ and let $\alpha$ and $\beta$ be constants. Then $\alpha f + \beta g$ is integrable on $[a, b]$ , and

\begin{aligned} &\int_a^b \bigl(\alpha f(x) + \beta g(x)\bigr)\,dx \\[0.4em] &\quad = \alpha \int_a^b f(x)\,dx \\[0.4em] &\qquad + \beta \int_a^b g(x)\,dx. \end{aligned}

This combines the sum rule and the constant multiple rule, and applying it repeatedly covers any finite number of terms. Together with the power rule it integrates every polynomial over every interval. In the language of linear algebra, the integral over a fixed interval is a linear map from the integrable functions on $[a, b]$ to the real numbers. The statement is about finitely many terms. Integrating an infinite series term by term needs further conditions.

In plain words Split the integrand at its plus and minus signs, integrate every term over the same interval, keep each constant with its own term, and add the results. Linearity covers sums and constant factors and nothing else. The integral of a product is in general not the product of the integrals, and the integral of a quotient is in general not a quotient of integrals, so read the operation that joins the pieces before splitting.

Example The terms $3x^2$ , $-4x$ and $2$ are polynomials, so each is continuous and hence integrable on $[0, 2]$ . Linearity and the power rule give

\begin{aligned} &\int_0^2 (3x^2 - 4x + 2)\,dx \\[0.4em] &\quad = 3\int_0^2 x^2\,dx \\[0.4em] &\qquad - 4\int_0^2 x\,dx \\[0.4em] &\qquad + 2\int_0^2 1\,dx \\[0.4em] &\quad = 3 \cdot \tfrac83 - 4 \cdot 2 + 2 \cdot 2 \\[0.4em] &\quad = 8 - 8 + 4 = 4. \end{aligned}

All three integrals keep the limits $0$ and $2$ . One antiderivative checks the whole calculation at once. The function $F(x) = x^3 - 2x^2 + 2x$ has $F'(x) = 3x^2 - 4x + 2$ , which confirms every coefficient and sign, and the integrand is continuous on $[0, 2]$ , so the fundamental theorem of calculus gives

\begin{aligned} F(2) - F(0) &= (8 - 8 + 4) - 0 \\[0.4em] &= 4. \end{aligned}

Signed contribution

Definition Write an integrand on $[a, b]$ , with $a < b$ , as a sum of finitely many integrable terms, $f = g_1 + \dots + g_n$ . The signed contribution of the term $g_k$ to $\int_a^b f(x)\,dx$ is its own integral $\int_a^b g_k(x)\,dx$ . It equals the area of the region between the graph of $g_k$ and the $x$ -axis where $g_k$ is positive, minus the area of the region where $g_k$ is negative. By linearity the signed contributions add up to the integral of $f$ . A negative contribution does not mean that $f$ is negative anywhere, since the value of $f$ at a point depends on all the terms at that point.

In plain words Each term moves the total up or down by its own signed area. A region below the axis has a positive size but makes a negative contribution, so note the size and the sign separately and carry the sign into the sum.

Example At $x = 1$ the terms of $3x^2 - 4x + 2$ have heights $3$ , $-4$ and $2$ , which add to the height $1$ of the combined graph. The term $-4x$ is negative on $(0, 2]$ . Its region is a triangle below the axis with base $2$ and depth $8$ , since $-4 \cdot 2 = -8$ , so its area is $\tfrac12 \cdot 2 \cdot 8 = 8$ and its signed contribution is

\begin{aligned} \int_0^2 (-4x)\,dx &= -4 \cdot 2 \\[0.4em] &= -8. \end{aligned}

With the contributions $8$ from $3x^2$ and $4$ from the constant $2$ , the total is $8 - 8 + 4 = 4$ . The contributions of $3x^2$ and $-4x$ cancel as numbers, but the two functions do not cancel point by point. Their sum $3x^2 - 4x = x(3x - 4)$ is zero on $[0, 2]$ only at $0$ and $\tfrac43$ , negative between them and positive after. The combined integrand never dips below the axis, because

\begin{aligned} &3x^2 - 4x + 2 \\[0.4em] &\quad = 3\left(x - \tfrac23\right)^2 + \tfrac23 \\[0.4em] &\quad \ge \tfrac23, \end{aligned}

with the smallest value $\tfrac23$ at $x = \tfrac23$ . So the value $4$ is also the area of the region under the combined graph. The negative contribution belongs to the split into terms, not to any part of the final graph.

Average value of a function

Definition Let $f$ be integrable on $[a, b]$ with $a < b$ . The average value of $f$ on $[a, b]$ is

f_{\text{avg}} = \frac{1}{b - a} \int_a^b f(x)\,dx.

It is the height of the rectangle on $[a, b]$ whose signed area equals the integral, and it carries the units of $f$ . Because $1/(b - a)$ is a constant, linearity carries over, and the average of $\alpha f + \beta g$ is $\alpha$ times the average of $f$ plus $\beta$ times the average of $g$ . If $f$ is continuous on $[a, b]$ , the mean value theorem for integrals gives at least one point $c$ in $(a, b)$ with $f(c) = f_{\text{avg}}$ . Without continuity the average may never be taken, as for the function that is $0$ on $[0, 1)$ and $1$ on $[1, 2]$ , whose average on $[0, 2]$ is $\tfrac12$ .

In plain words Flatten the region into a rectangle on the same base with the same signed area, and the height of that rectangle is the average value. It averages the function over every point of the interval, not just its two endpoint values. Where the graph runs below the average height, the missing area exactly matches the extra area where it runs above.

Example For $f(x) = 3x^2 - 4x + 2$ on $[0, 2]$ ,

\begin{aligned} f_{\text{avg}} &= \frac{1}{2 - 0} \int_0^2 f(x)\,dx \\[0.4em] &= \frac{4}{2} = 2. \end{aligned}

By linearity the averages $4$ , $-4$ and $2$ of the three terms add to the same $2$ . The graph is far from constant at height $2$ . It starts at $f(0) = 2$ , falls to $\tfrac23$ at $x = \tfrac23$ and climbs to $f(2) = 6$ . It meets the height $2$ where $3x^2 - 4x = 0$ , at $x = 0$ and $x = \tfrac43$ , so $c = \tfrac43$ is a point of $(0, 2)$ where $f$ takes its average value. The area missing below the line $y = 2$ on $[0, \tfrac43]$ equals the extra area above it on $[\tfrac43, 2]$ ,

\begin{aligned} \int_0^{4/3} (4x - 3x^2)\,dx &= \tfrac{32}{27}, \\[0.4em] \int_{4/3}^2 (3x^2 - 4x)\,dx &= \tfrac{32}{27}. \end{aligned}

The average of the two endpoint values, $\tfrac12(2 + 6) = 4$ , is twice the true average.

Check yourself

Use linearity to evaluate the integral below, then check the result with one antiderivative.

\int_1^3 (6x^2 - 2x + 5)\,dx

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The three terms are continuous on $[1, 3]$ . The power rule gives $\int_1^3 x^2\,dx = \tfrac{27 - 1}{3} = \tfrac{26}{3}$ and $\int_1^3 x\,dx = \tfrac{9 - 1}{2} = 4$ , and the constant $1$ integrates to the width $2$ . By linearity,

\begin{aligned} &\int_1^3 (6x^2 - 2x + 5)\,dx \\[0.4em] &\quad = 6 \cdot \tfrac{26}{3} - 2 \cdot 4 + 5 \cdot 2 \\[0.4em] &\quad = 52 - 8 + 10 = 54. \end{aligned}

The function $F(x) = 2x^3 - x^2 + 5x$ has $F'(x) = 6x^2 - 2x + 5$ , and $F(3) - F(1) = 60 - 6 = 54$ . The lower limit is $1$ , not $0$ , so the value $F(1) = 6$ must be subtracted.

Let $g(x) = 2x^2 - 6x + 4$ . Evaluate $\int_0^3 g(x)\,dx$ by linearity. Is the result the area of the region between the graph of $g$ and the $x$ -axis over $[0, 3]$ ?

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Since $\int_0^3 x^2\,dx = 9$ and $\int_0^3 x\,dx = \tfrac92$ , the signed contributions are $18$ , $-27$ and $12$ , and

\begin{aligned} &\int_0^3 (2x^2 - 6x + 4)\,dx \\[0.4em] &\quad = 2 \cdot 9 - 6 \cdot \tfrac92 + 4 \cdot 3 \\[0.4em] &\quad = 18 - 27 + 12 = 3. \end{aligned}

This is not the area. The integrand factors as $g(x) = 2(x - 1)(x - 2)$ , which is negative for $1 < x < 2$ , so this time the combined graph itself dips below the axis. With the antiderivative $G(x) = \tfrac23 x^3 - 3x^2 + 4x$ , the values $G(0) = 0$ , $G(1) = \tfrac53$ , $G(2) = \tfrac43$ and $G(3) = 3$ give the pieces

\begin{aligned} \int_0^1 g(x)\,dx &= \tfrac53, \\[0.4em] \int_1^2 g(x)\,dx &= -\tfrac13, \\[0.4em] \int_2^3 g(x)\,dx &= \tfrac53. \end{aligned}

The integral $3$ is the area above the axis minus the area below it, while the area of the region is $\tfrac53 + \tfrac13 + \tfrac53 = \tfrac{11}{3}$ . The question is decided by the sign of $g$ itself, not by the signs of its terms.

Let $h(x) = x(x + 1)$ . A classmate computes $\int_0^2 h(x)\,dx$ as the product of $\int_0^2 x\,dx$ and $\int_0^2 (x + 1)\,dx$ . Explain the mistake, find the correct integral, and find the average value of $h$ on $[0, 2]$ .

Show the answer

Linearity splits sums and moves constant factors outside. It says nothing about products. The two integrals are $\int_0^2 x\,dx = 2$ and $\int_0^2 (x + 1)\,dx = 4$ , so the classmate gets $8$ , which is not the integral of $h$ . Expanding first turns the product into a sum, $h(x) = x^2 + x$ , and then

\begin{aligned} \int_0^2 (x^2 + x)\,dx &= \tfrac83 + 2 \\[0.4em] &= \tfrac{14}{3}. \end{aligned}

The average value is

h_{\text{avg}} = \frac{1}{2} \cdot \frac{14}{3} = \frac{7}{3}.

The function $h$ is continuous, so it takes this value somewhere in $(0, 2)$ . Solving $x^2 + x = \tfrac73$ gives $x = (-3 + \sqrt{93})/6 \approx 1.107$ as the root in that interval. The average of the endpoint values, $\tfrac12(0 + 6) = 3$ , is a different number and is not the average value.

Linearity and signed contributions for a sum of terms

Key terms

Sum rule for integrals

Constant multiple rule

Linearity of the integral

Signed contribution

Average value of a function

Check yourself

How this lesson connects

Builds on

Used later in