Unit 6: Integration and Accumulation of Change

Suppose a tank's net flow-rate graph, in liters per minute, forms a triangle above the axis with base $4$ minutes and height $6$ liters per minute, followed by a triangle below the axis with base $2$ minutes and height $3$ liters per minute. The net change is $\tfrac12(4)(6)-\tfrac12(2)(3)=12-3=9$ liters. If the tank began with $20$ liters, it ends with $29$ liters.

Suppose a table gives $x=0,1,3,4$ and $f(x)=2,4,5,3$. The widths are $1,2,1$, not all equal. The left sum is $2(1)+4(2)+5(1)=15$. The right sum is $4(1)+5(2)+3(1)=17$. The trapezoidal sum is $\tfrac{2+4}{2}(1)+\tfrac{4+5}{2}(2)+\tfrac{5+3}{2}(1)=16$.

Refining the partition makes every rectangle narrower, so the sum follows the changing function more closely. The condition $\max \Delta x_i\to0$ matters: it prevents even one subinterval from remaining wide. For a continuous function, different valid choices of sample points approach the same definite integral.

For a positive rate $r(t)$ on $[0,4]$, a sum $\sum r(t_i^*)\Delta t_i$ adds approximate changes over short time intervals. As the largest time width approaches zero, the approximation approaches $\int_0^4 r(t)\,dt$, the exact accumulated change over those four time units.

In $\lim_{n\to\infty}\sum_{i=1}^n(1+3i/n)^2(3/n)$, the width is $3/n$, so $b-a=3$. The sample point is $1+3i/n$, so $a=1$ and $b=4$. The squared factor is $f(x_i)$. Therefore the limit is $\int_1^4 x^2\,dx$.

For $\int_0^2\sqrt{1+x^3}\,dx$, use $\Delta x=2/n$ and right endpoints $x_i=2i/n$. An equivalent limit is $\lim_{n\to\infty}\sum_{i=1}^n\sqrt{1+(2i/n)^3}(2/n)$. The factor outside the function is the subinterval width.

If $f$ is continuous on an interval containing $a$ and $x$, increasing the endpoint by a small amount $\Delta x$ adds approximately $f(x)\Delta x$ to the accumulation. Dividing by $\Delta x$ and taking the limit leaves $f(x)$. This is the Fundamental Theorem connection between integration and differentiation.

Let $A(x)=\int_0^x(2t+1)\,dt$. Then $A(0)=0$, because no interval has yet been accumulated. The Fundamental Theorem gives $A'(x)=2x+1$. Evaluating directly, $A(x)=x^2+x$, so $A(2)=6$ and differentiating this expression confirms the same derivative.

Suppose the signed area under $f$ from $a$ to $c$ is $5$, and the area from $c$ to $d$ lies below the axis with magnitude $2$. Then $g(c)=5$ and $g(d)=5-2=3$. On $(c,d)$, $f<0$, so $g$ is decreasing, yet $g$ remains positive. If $f$ changes from positive to negative at $c$, then $g$ has a local maximum there.

If $\int_0^2 f(x)\,dx=5$ and $\int_2^7 f(x)\,dx=-1$, then $\int_0^7 f(x)\,dx=5+(-1)=4$. Therefore $\int_7^0 3f(x)\,dx=3[-\int_0^7 f(x)\,dx]=3(-4)=-12$. The negative result comes from reversing the limits, not from discarding the given signed value.

If a graph on $[-2,2]$ is the upper semicircle of radius $2$, its integral is the semicircle's area, $\tfrac12\pi(2)^2=2\pi$. The same semicircle below the axis would contribute $-2\pi$.

Evaluate $\int_1^3(2x^2-4x+1)\,dx$. A polynomial is continuous, and one antiderivative is $F(x)=\tfrac23x^3-2x^2+x$. Then $F(3)=3$ and $F(1)=-\tfrac13$, so $\int_1^3(2x^2-4x+1)\,dx=3-(-\tfrac13)=\tfrac{10}{3}$.

For $x\ne0$, $\int(6x^2-4/x+3e^x)\,dx=2x^3-4\ln|x|+3e^x+C$. Differentiating gives $6x^2-4/x+3e^x$, exactly the original integrand, so the coefficients, signs, and logarithm condition are consistent.

For $\int 6x(3x^2+1)^4\,dx$, let $u=3x^2+1$, so $du=6x\,dx$. Then $\int u^4\,du=u^5/5+C$, giving $(3x^2+1)^5/5+C$. Differentiation returns the original integrand.

For $\int_0^1 2x e^{x^2}\,dx$, let $u=x^2$ and $du=2x\,dx$. The original bounds become $u(0)=0$ and $u(1)=1$, so the integral is $\int_0^1 e^u\,du=[e^u]_0^1=e-1$.

For $x\ne-1$, $\frac{x^2+1}{x+1}=x-1+\frac{2}{x+1}$. Therefore $\int\frac{x^2+1}{x+1}\,dx=\frac{x^2}{2}-x+2\ln|x+1|+C$. Multiplying the quotient by the divisor and adding the remainder verifies the rewrite.

Since $x^2+4x+8=(x+2)^2+4$, $\int_{-2}^{0}\frac{dx}{x^2+4x+8}=\left[\frac12\arctan\left(\frac{x+2}{2}\right)\right]_{-2}^{0}=\frac12(\pi/4-0)=\pi/8$. The completed square identifies the inverse-tangent form.

For $\int xe^x\,dx$, take $u=x$ and $dv=e^x\,dx$. Then $du=dx$ and $v=e^x$, so $\int xe^x\,dx=xe^x-\int e^x\,dx=xe^x-e^x+C$. Differentiating the result returns $xe^x$.

For a definite integral, keep the bounds: $\int_0^1 xe^x\,dx=[xe^x]_0^1-\int_0^1e^x\,dx=[xe^x-e^x]_0^1=0-(-1)=1$. The result is a number, so no $+C$ is added.

Decompose $\frac{3x+5}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}$. Multiplying through gives $3x+5=A(x+2)+B(x+1)$, so $A=2$ and $B=1$. Therefore $\int\frac{3x+5}{(x+1)(x+2)}\,dx=2\ln|x+1|+\ln|x+2|+C$.

Since $\frac{1}{(x+1)(x+2)}=\frac{1}{x+1}-\frac{1}{x+2}$, $\int_0^1\frac{dx}{(x+1)(x+2)}=[\ln(x+1)-\ln(x+2)]_0^1=\ln(4/3)$. The interval contains no denominator zero, so the integrand is continuous there.

For an infinite interval, $\int_1^{\infty}x^{-2}\,dx=\lim_{b\to\infty}[-x^{-1}]_1^b=\lim_{b\to\infty}(1-1/b)=1$. The finite limit means the integral converges to $1$.

The integrand $1/x^2$ is unbounded at $0$, so $\int_{-1}^{1}x^{-2}\,dx$ must be split. On the right, $\lim_{t\to0^+}\int_t^1x^{-2}\,dx=\lim_{t\to0^+}(1/t-1)=\infty$. Because one required one-sided integral diverges, the original integral diverges.

Examples of selection: $\int 2x\cos(x^2)\,dx$ suggests $u=x^2$ because $du=2x\,dx$ is present. In contrast, $\int xe^x\,dx$ has no inner-function pattern, but differentiating $x$ simplifies it, so integration by parts is appropriate.

For $\int\frac{x^2+1}{x+1}\,dx$, the numerator degree signals long division before integrating. For $\int\frac{1}{(x+1)(x+2)}\,dx$, the proper rational form with distinct linear factors signals partial fractions. These decisions come from structure, not from the Topic number.