Fundamentals and Basic Rules
Core antiderivative rules, definite-integral properties, and the Fundamental Theorem of Calculus.
Core antiderivative rules, definite-integral properties, and the Fundamental Theorem of Calculus.
Evaluate $$\int_2^7 \frac{3x}{2x^2+4}\,dx.$$
Let $u=2x^2+4$, so $du=4x\,dx$.
$$\int_2^7 \frac{3x}{2x^2+4}\,dx=\frac{3}{4}\int_{12}^{102}\frac{du}{u}=\frac{3}{4}\ln\left(\frac{102}{12}\right)$$
$$\boxed{1.605}$$
Evaluate $$\int_2^5 \frac{x+1}{x^2-x}\,dx.$$
Decompose the rational expression:
$$\frac{x+1}{x(x-1)}=-\frac{1}{x}+\frac{2}{x-1}$$
Then
$$\int_2^5\left(-\frac{1}{x}+\frac{2}{x-1}\right)dx=\left[-\ln x+2\ln|x-1|\right]_2^5$$
$$=\ln\left(\frac{32}{5}\right)=\boxed{1.856}$$
Find $y=f(x)$ if the graph passes through $(1,2)$ and satisfies $$\frac{dy}{dx}=3x^2-3.$$
$$y=\int(3x^2-3)\,dx=x^3-3x+C$$
Use $(1,2)$:
$$2=1-3+C\quad\Rightarrow\quad C=4$$
$$\boxed{y=x^3-3x+4}$$
If $f(2)=4$ and $$f(x)=\int(3x^3-7x)\,dx,$$ find $f(0)$.
$$f(x)=\frac{3}{4}x^4-\frac{7}{2}x^2+C$$
Use $f(2)=4$:
$$4=12-14+C\quad\Rightarrow\quad C=6$$
Therefore, $$\boxed{f(0)=6}$$
Evaluate $$\int_0^9 \frac{dx}{1+\sqrt{x}}.$$
Let $u=\sqrt{x}$, so $x=u^2$ and $dx=2u\,du$. The limits become $u=0$ to $u=3$.
$$\int_0^9 \frac{dx}{1+\sqrt{x}}=\int_0^3\frac{2u}{1+u}\,du=\int_0^3\left(2-\frac{2}{1+u}\right)du$$
$$=\left[2u-2\ln(1+u)\right]_0^3=6-2\ln4=\boxed{3.227}$$
Without antiderivatives, evaluate $\displaystyle\int_{-3}^{3}\sqrt{9-x^2}\,dx$ using geometry.
The integrand $y=\sqrt{9-x^2}$ is the upper half of the circle $x^2+y^2=9$ (radius 3). The integral from $-3$ to $3$ sweeps the entire upper semicircle.
This is a powerful shortcut on board exams — always check if the integrand is a recognizable geometric shape before computing.
Evaluate $\displaystyle\int_{-2}^{2}(x^3+2x)\,dx$ using properties of odd functions, without computing directly.
Let $f(x)=x^3+2x$. Check: $f(-x)=(-x)^3+2(-x)=-x^3-2x=-(x^3+2x)=-f(x)$.
Since $f$ is an odd function and the interval is symmetric about 0:
The area above the axis exactly cancels the area below. This is a 5-second problem once you recognize the pattern.
Find $\dfrac{d}{dx}\displaystyle\int_1^{x^2}\sin(t^3)\,dt$.
By the Fundamental Theorem of Calculus Part 2, $\dfrac{d}{dx}\int_a^{g(x)}f(t)\,dt=f(g(x))\cdot g'(x)$.
Here $g(x)=x^2$, so $g'(x)=2x$, and $f(t)=\sin(t^3)$.
Evaluate $$\int (8x^3 - 2x + \sqrt{5})\,dx.$$
Use linearity: integrate each term separately. The constant $\sqrt{5}$ is treated like any number.
$$\int 8x^3\,dx=8\left(\frac{x^4}{4}\right)=2x^4$$
$$\int -2x\,dx=-2\left(\frac{x^2}{2}\right)=-x^2$$
$$\int \sqrt{5}\,dx=\sqrt{5}\,x$$
Combine the terms and add the constant of integration:
$$\boxed{2x^4-x^2+\sqrt{5}\,x+C}$$
Evaluate $$\int (3 - 2x)(4x + 5)\,dx.$$
First expand the product so the power rule can be applied.
$$(3-2x)(4x+5)=12x+15-8x^2-10x=-8x^2+2x+15$$
Now integrate term by term:
$$\int(-8x^2+2x+15)\,dx=-8\left(\frac{x^3}{3}\right)+2\left(\frac{x^2}{2}\right)+15x+C$$
$$\boxed{-\frac{8}{3}x^3+x^2+15x+C}$$
Evaluate $$\int \frac{(x - 2)^3}{3x^2}\,dx.$$
Expand the numerator first:
$$(x-2)^3=x^3-6x^2+12x-8$$
Divide every term by $3x^2$:
$$\frac{x^3-6x^2+12x-8}{3x^2}=\frac{x}{3}-2+\frac{4}{x}-\frac{8}{3x^2}$$
Now integrate. Remember that $\int \frac{1}{x}\,dx=\ln|x|$.
$$\int\left(\frac{x}{3}-2+\frac{4}{x}-\frac{8}{3}x^{-2}\right)dx$$
$$=\frac{x^2}{6}-2x+4\ln|x|+\frac{8}{3x}+C$$
$$\boxed{\frac{x^2}{6}-2x+4\ln|x|+\frac{8}{3x}+C}$$
Evaluate $$\int (x^{3/4} + 2x^{1/2})^2\,dx.$$
Use $(a+b)^2=a^2+2ab+b^2$.
$$(x^{3/4}+2x^{1/2})^2=x^{3/2}+4x^{5/4}+4x$$
Apply the power rule $\int x^n\,dx=\frac{x^{n+1}}{n+1}+C$.
$$\int x^{3/2}dx=\frac{2}{5}x^{5/2}$$
$$\int 4x^{5/4}dx=4\left(\frac{x^{9/4}}{9/4}\right)=\frac{16}{9}x^{9/4}$$
$$\int 4x\,dx=2x^2$$
$$\boxed{\frac{2}{5}x^{5/2}+\frac{16}{9}x^{9/4}+2x^2+C}$$
Evaluate $$\int \sqrt{x^{1/2} - 2x^{3/2} + x^{5/2}}\,dx.$$
Factor the expression inside the radical.
$$x^{1/2}-2x^{3/2}+x^{5/2}=x^{1/2}(1-2x+x^2)=x^{1/2}(x-1)^2$$
Take the square root carefully:
$$\sqrt{x^{1/2}(x-1)^2}=x^{1/4}|x-1|$$
Because of the absolute value, the antiderivative depends on whether $x\ge1$ or $0\le x\le1$.
For $x\ge1$, $|x-1|=x-1$:
$$\int x^{1/4}(x-1)dx=\int(x^{5/4}-x^{1/4})dx=\frac{4}{9}x^{9/4}-\frac{4}{5}x^{5/4}+C$$
For $0\le x\le1$, $|x-1|=1-x$:
$$\int x^{1/4}(1-x)dx=\int(x^{1/4}-x^{5/4})dx=\frac{4}{5}x^{5/4}-\frac{4}{9}x^{9/4}+C$$
$$ \boxed{ \begin{cases} \frac{4}{9}x^{9/4}-\frac{4}{5}x^{5/4}+C, & x\ge 1\\ \frac{4}{5}x^{5/4}-\frac{4}{9}x^{9/4}+C, & 0\le x\le 1 \end{cases}} $$
Evaluate $$\int \left(\frac{4}{x^2} - \frac{3}{x^3} + \frac{2}{x^4}\right)\,dx.$$
Rewrite the fractions using negative exponents.
$$\frac{4}{x^2}-\frac{3}{x^3}+\frac{2}{x^4}=4x^{-2}-3x^{-3}+2x^{-4}$$
Integrate each power:
$$\int4x^{-2}dx=-4x^{-1}=-\frac{4}{x}$$
$$\int-3x^{-3}dx=\frac{3}{2}x^{-2}=\frac{3}{2x^2}$$
$$\int2x^{-4}dx=-\frac{2}{3}x^{-3}=-\frac{2}{3x^3}$$
$$\boxed{-\frac{4}{x}+\frac{3}{2x^2}-\frac{2}{3x^3}+C}$$
Evaluate $$\int \sqrt[3]{t}(\sqrt{t} + 1)^2\,dt.$$
Convert radicals to fractional exponents: $\sqrt[3]{t}=t^{1/3}$ and $\sqrt{t}=t^{1/2}$.
First expand the square:
$$(\sqrt{t}+1)^2=t+2t^{1/2}+1$$
Multiply by $t^{1/3}$:
$$t^{1/3}(t+2t^{1/2}+1)=t^{4/3}+2t^{5/6}+t^{1/3}$$
Integrate term by term:
$$\boxed{\frac{3}{7}t^{7/3}+\frac{12}{11}t^{11/6}+\frac{3}{4}t^{4/3}+C}$$
Evaluate $$\int \frac{(5 - 2\sqrt{t})^2}{\sqrt[4]{t^3}}\,dt.$$
Rewrite the denominator as $t^{3/4}$ and expand the numerator.
$$(5-2\sqrt{t})^2=25-20t^{1/2}+4t$$
Divide each term by $t^{3/4}$:
$$25t^{-3/4}-20t^{-1/4}+4t^{1/4}$$
Now use the power rule:
$$\int25t^{-3/4}dt=100t^{1/4}$$
$$\int-20t^{-1/4}dt=-\frac{80}{3}t^{3/4}$$
$$\int4t^{1/4}dt=\frac{16}{5}t^{5/4}$$
$$\boxed{100t^{1/4}-\frac{80}{3}t^{3/4}+\frac{16}{5}t^{5/4}+C}$$
Evaluate $$\int \frac{x^{4/3} - 16}{x^{1/3} + 2}\,dx.$$
Let $u=x^{1/3}$ only to simplify the algebraic division. Then $x^{4/3}=u^4$.
$$\frac{x^{4/3}-16}{x^{1/3}+2}=\frac{u^4-16}{u+2}$$
Factor or divide:
$$\frac{u^4-16}{u+2}=u^3-2u^2+4u-8$$
Return to $x$:
$$x-2x^{2/3}+4x^{1/3}-8$$
Integrate each term:
$$\int(x-2x^{2/3}+4x^{1/3}-8)dx$$
$$=\frac{x^2}{2}-\frac{6}{5}x^{5/3}+3x^{4/3}-8x+C$$
$$\boxed{\frac{x^2}{2}-\frac{6}{5}x^{5/3}+3x^{4/3}-8x+C}$$
Evaluate $$\int \left(4 - \frac{3}{\sqrt{x}}\right)^2 \frac{dx}{\sqrt{x}}.$$
Convert $\sqrt{x}$ to $x^{1/2}$ and expand the square.
$$\left(4-\frac{3}{\sqrt{x}}\right)^2=16-\frac{24}{\sqrt{x}}+\frac{9}{x}$$
Now multiply by $\frac{1}{\sqrt{x}}=x^{-1/2}$:
$$16x^{-1/2}-24x^{-1}+9x^{-3/2}$$
Integrate each term. The middle term gives a logarithm:
$$\int16x^{-1/2}dx=32\sqrt{x}$$
$$\int-24x^{-1}dx=-24\ln|x|$$
$$\int9x^{-3/2}dx=-18x^{-1/2}=-\frac{18}{\sqrt{x}}$$
$$\boxed{32\sqrt{x}-24\ln|x|-\frac{18}{\sqrt{x}}+C}$$
Evaluate $$\int (8x^3 - 3x^2 + 5x + 3)\,dx.$$
This is a direct power-rule problem.
$$\int8x^3dx=2x^4$$
$$\int-3x^2dx=-x^3$$
$$\int5x\,dx=\frac{5}{2}x^2$$
$$\int3\,dx=3x$$
$$\boxed{2x^4-x^3+\frac{5}{2}x^2+3x+C}$$
Evaluate $$\int_{2}^{3} \left(1 - \frac{x}{2}\right)^2\,dx.$$
Expand the square first:
$$\left(1-\frac{x}{2}\right)^2=1-x+\frac{x^2}{4}$$
Find an antiderivative:
$$\int\left(1-x+\frac{x^2}{4}\right)dx=x-\frac{x^2}{2}+\frac{x^3}{12}$$
Evaluate from $2$ to $3$:
$$\left[x-\frac{x^2}{2}+\frac{x^3}{12}\right]_2^3=\left(3-\frac{9}{2}+\frac{27}{12}\right)-\left(2-\frac{4}{2}+\frac{8}{12}\right)$$
$$=\frac{3}{4}-\frac{2}{3}=\boxed{\frac{1}{12}}$$
Evaluate $$\int_{2}^{8} \frac{dy}{y\sqrt{2y}}.$$
Rewrite the denominator using exponents:
$$y\sqrt{2y}=y\sqrt{2}\sqrt{y}=\sqrt{2}\,y^{3/2}$$
So the integrand is
$$\frac{1}{y\sqrt{2y}}=\frac{1}{\sqrt{2}}y^{-3/2}$$
Integrate:
$$\int\frac{1}{\sqrt{2}}y^{-3/2}dy=\frac{1}{\sqrt{2}}\left(-2y^{-1/2}\right)=-\frac{\sqrt{2}}{\sqrt{y}}$$
Evaluate from $2$ to $8$:
$$\left[-\frac{\sqrt{2}}{\sqrt{y}}\right]_2^8=-\frac{\sqrt{2}}{\sqrt{8}}-\left(-\frac{\sqrt{2}}{\sqrt{2}}\right)=-\frac{1}{2}+1$$
$$\boxed{\frac{1}{2}}$$
Evaluate $$\int_{1}^{9} \sqrt{x(4 - x)}\,dx.$$
Before integrating, always check the domain of an even root. For a real square root, the radicand must be nonnegative:
$$x(4-x)\ge0$$
This is true only for $0\le x\le4$. But the interval $[1,9]$ includes values greater than $4$.
For example, at $x=5$:
$$x(4-x)=5(4-5)=-5$$
Since $\sqrt{-5}$ is not real, the integrand is not real-valued on the whole interval $[1,9]$.
Therefore, as written, the real-valued definite integral is not defined over $[1,9]$.
$$\boxed{\text{No real-valued definite integral as written.}}$$
Evaluate $$\int_{1}^{4} \left(4x - \frac{5}{x\sqrt{x}}\right)\,dx.$$
Rewrite the radical part using exponents:
$$x\sqrt{x}=x\cdot x^{1/2}=x^{3/2}$$
So
$$4x-\frac{5}{x\sqrt{x}}=4x-5x^{-3/2}$$
Find an antiderivative:
$$\int(4x-5x^{-3/2})dx=2x^2+10x^{-1/2}=2x^2+\frac{10}{\sqrt{x}}$$
Evaluate from $1$ to $4$:
$$\left[2x^2+\frac{10}{\sqrt{x}}\right]_1^4=\left(2(4)^2+\frac{10}{2}\right)-\left(2(1)^2+\frac{10}{1}\right)$$
$$=(32+5)-(2+10)=37-12=\boxed{25}$$