Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

H(x)=-x-5 h(x)=H^(')(x) int_(-3)^(6)h(x)dx=

H(x)=−x−5 H(x)=-x-5 \newlineh(x)=H′(x) h(x)=H^{\prime}(x) \newline∫−36h(x)dx= \int_{-3}^{6} h(x) d x=

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Calculusright chevron icon
Evaluate definite integrals using the power ruleright chevron icon

Full solution

Q. H(x)=−x−5 H(x)=-x-5 \newlineh(x)=H′(x) h(x)=H^{\prime}(x) \newline∫−36h(x)dx= \int_{-3}^{6} h(x) d x=
  1. Integrate h(x)h(x): Now we integrate h(x)h(x) from −3-3 to 66.∫−36h(x) dx=∫−36(−1) dx\int_{-3}^{6} h(x) \, dx = \int_{-3}^{6} (-1) \, dxThis is a simple integral, the antiderivative of −1-1 is −x-x.
  2. Evaluate antiderivative: We evaluate the antiderivative at the bounds 66 and −3-3.−(−6)−(−(−3))=6−3=3-(-6) - (-(-3)) = 6 - 3 = 3
  3. Correct evaluation: But wait, there's a mistake in the previous step. The correct evaluation should be:\newline−(−6)−(−(−3))=6+3=9-(-6) - (-(-3)) = 6 + 3 = 9

More problems from Evaluate definite integrals using the power rule

Question
The number of subscribers to a magazine is changing at a rate of r(t) r(t) subscribers per month (where t t is time in months).\newlineWhat does ∫810r′(t)dt=7 \int_{8}^{10} r^{\prime}(t) d t=7 mean?\newlineChoose 11 answer:\newline(A) The rate of change of number of subscribers increased by 77 subscribers per month between t=8 t=8 and t=10 t=10 months.\newline(B) As of month 1010, the magazine had 77 subscribers.\newline(C) The number of subscribers increased by 77 between t=8 t=8 and t=10 t=10 months.\newline(D) The average rate of change in subscribers between month 88 and month 1010 was 77 subscribers per month.
Get tutor helpright-arrow

Posted 3 months ago

Question
Consider the following problem:\newlineThe water level under a bridge is changing at a rate of r(t)=40sin⁡(πt6) r(t)=40 \sin \left(\frac{\pi t}{6}\right) centimeters per hour (where t t is the time in hours). At time t=3 t=3 , the water level is 9191 centimeters. By how much does the water level change during the 4th  4^{\text {th }} hour?\newlineWhich expression can we use to solve the problem?\newlineChoose 11 answer:\newline(A) ∫04r(t)dt \int_{0}^{4} r(t) d t \newline(B) ∫45r(t)dt \int_{4}^{5} r(t) d t \newline(C) ∫34r(t)dt \int_{3}^{4} r(t) d t \newline(D) ∫44r(t)dt \int_{4}^{4} r(t) d t
Get tutor helpright-arrow

Posted 3 months ago

Question
The base of a solid is the region enclosed by the graphs of y=sin⁡(x) y=\sin (x) and y=4−x y=4-\sqrt{x} , between x=2 x=2 and x=7 x=7 .\newlineCross sections of the solid perpendicular to the x x -axis are rectangles whose height is 2x 2 x .\newlineWhich one of the definite integrals gives the volume of the solid?\newlineChoose 11 answer:\newline(A) ∫27[sin⁡(x)+x−4]⋅2xdx \int_{2}^{7}[\sin (x)+\sqrt{x}-4] \cdot 2 x d x \newline(B) ∫27[sin⁡2(x)−(4−x)2]dx \int_{2}^{7}\left[\sin ^{2}(x)-(4-\sqrt{x})^{2}\right] d x \newline(C) ∫27[4−x−sin⁡(x)]⋅2xdx \int_{2}^{7}[4-\sqrt{x}-\sin (x)] \cdot 2 x d x \newline(D) ∫27[(4−x)2−sin⁡2(x)]dx \int_{2}^{7}\left[(4-\sqrt{x})^{2}-\sin ^{2}(x)\right] d x
Get tutor helpright-arrow

Posted 3 months ago

Question
Nork/en/geometry/unit/99/lesson/11\newlineCoursework\newlineEdgext\newlineWall\newlineMenu\newlineTriangle \newline△OGK \triangle OGK is shown where \newlineOG=22.3 OG=22.3 centimeters (cond \newlineGK=32.5cm GK=32.5\text{cm} , OK=21.9cm OK=21.9\text{cm} . Height of the triangle is \newline15cm 15\text{cm} .\newlineWhat is the area of \newline△OGK \triangle OGK ?\newline(A) \newline244.185cm2 244.185\text{cm}^2 \newline(B) \newline243.75cm2 243.75\text{cm}^2 \newline(C) \newline167.25cm2 167.25\text{cm}^2 \newline(D) \newline487.5cm2 487.5\text{cm}^2
Get tutor helpright-arrow

Posted 2 months ago

Question
∫(ecot⁡−1(2z))14+z2dz\int \left(e^{\cot^{-1}\left(\frac{2}{z}\right)}\right)\frac{1}{4+z^{2}}dz
Get tutor helpright-arrow

Posted 2 months ago

Question
∫0π(1−cos⁡(x)) dx=\int_{0}^{\pi}(1-\cos(x))\,dx=
Get tutor helpright-arrow

Posted 2 months ago

Question
∫(1+3x)x2 dx\int (1+3x)x^{2}\,dx
Get tutor helpright-arrow

Posted 2 months ago

Question
∫sin⁡6(x)cos⁡3(x) dx\int \frac{\sin^6(x)}{\cos^3(x)} \, dx
Get tutor helpright-arrow

Posted 2 months ago

Question
Find ∫(2+5x−1(x−2)2)dx \int(2+5x-\frac{1}{(x-2)^{2}})dx
Get tutor helpright-arrow

Posted 2 months ago

Question
Consider the reaction between Na2CO3(aq)\text{Na}_2\text{CO}_3(\text{aq}) and H3PO4(aq)\text{H}_3\text{PO}_4(\text{aq}) as shown in the following balanced equation:\newline3Na2CO3(aq)+2H3PO4(aq)→2Na3PO4(aq)+3H2O(l)+3CO2(g)3\text{Na}_2\text{CO}_3(\text{aq})+2\text{H}_3\text{PO}_4(\text{aq})\rightarrow 2\text{Na}_3\text{PO}_4(\text{aq})+3\text{H}_2\text{O}(\text{l})+3\text{CO}_2(\text{g})\newline\newlineChemical\newlineMW\newline\newlineNa2CO3\text{Na}_2\text{CO}_3\newline105.991105.991\newline\newlineH3PO4\text{H}_3\text{PO}_4\newline97.997797.9977\newline\newlineNa3PO4\text{Na}_3\text{PO}_4\newline163.944163.944\newline\newlineH2O\text{H}_2\text{O}\newline18.015818.0158\newline\newlineH3PO4(aq)\text{H}_3\text{PO}_4(\text{aq})00\newlineH3PO4(aq)\text{H}_3\text{PO}_4(\text{aq})11\newline\newlineHow many gram of Na2CO3\text{Na}_2\text{CO}_3 is used if H3PO4(aq)\text{H}_3\text{PO}_4(\text{aq})33 is made assuming all reactant is converted into products?
Get tutor helpright-arrow

Posted 2 months ago

View all (26)